Important Integrals Questions for CBSE Class 12th

Let α ∈ 0 , π 2 be fixed. If ∫ sin ⁡ ( x + α ) sin ⁡ ( x − α ) d x = f 1 ( x ) cos ⁡ 2 α + f 2 ( x ) sin ⁡ 2 α + c where c constant of integration then

∫ { 1 + 2 tan ⁡ x ( tan ⁡ x + sec ⁡ x ) } 1 / 2 d x =

∫ cos ⁡ 2 x cos ⁡ x d x is equal to

The value of ∫ a b | x | x d x is

∫ f ( x ) d x = g ( x ) , t h e n ∫ f – 1 ( x ) d x =

Evaluate ∫ e x e 2 x + 6 e x + 5 d x

Evaluate ∫ 2 x − 1 ( x − 1 ) ( x + 2 ) ( x − 3 ) d x

∫ cos e c 4 x d x =

Evaluate ∫ 0 ∞ tan − 1 ⁡ x x 1 + x 2 d x

A curve g ( x ) = ∫ x 27 1 + x + x 2 6 6 x 2 + 5 x + 4 d x is passing through (0,0) then

∫ tan 2 x sec 4 x d x

∫ cot 2 x cos e c 4 x d x

∫ x 6 + x 4 + x 2 2 x 4 + 3 x 2 + 6 d x

∫ 2 x + tan − 1 x 1 + x 2 d x =

∫ ( sec ⁡ x + tan ⁡ x ) 2 d x =

∫ e x cos ⁡ e x x d x =

If I = ∫ d x tan ⁡ x log ⁡ cosec ⁡ x , then I equals

If f ( x ) = ∫ 5 x 8 + 7 x 6 x 2 + 1 + 2 x 7 2 d x , and f ( 0 ) = 0 , then the value of f ( 1 ) is

If f ‘ ( x ) = tan – 1 ( sec x + tan x ) , – π 2 < x < π 2 , and f ( 0 ) = 0 , then f ( 1 ) is equal to:

∫ d x ( a 2 + x 2 ) 3 2 =

∫ ( 1 + x ) 3 x d x =

Evaluate ∫ tan 4 ⁡ x d x

Evaluate ∫ 1 sin 4 ⁡ x + cos 4 ⁡ x d x

Evaluate ∫ e x 1 − sin ⁡ x 1 − cos ⁡ x d x

Evaluate ∫ sec 2 ⁡ x tan 2 ⁡ x + 4 d x

∫ x ( x – 1 ) ( x 2 + 4 ) d x

E v a l u a t e ∫ sin x sin 4 x d x

If I n = ∫ sin n ⁡ x d x then I 8 − 7 8 I 6 is

∫ sec 4 ⁡ x cosec 2 ⁡ x d x , is equal to

∫ x sin − 1 x 1 − x 2 d x =

if 2 < x < 3 ∫ ( | x − 1 | | x − 2 | + | x − 3 | ) d x

∫ cos x − c o x 2 x 1 − cos x d x =

∫ 1 1 + cos x d x =

∫ x e x 1 + x 2 d x =

∫ d x sin 3 x cos 5 x =

∫ tan ( x − α ) tan ( x + α ) tan 2 x d x =

∫ x 6 2 x + x 7 5 d x =

∫ log ⁡ ( log ⁡ ( log ⁡ x ) ) x log ⁡ x log ⁡ ( log ⁡ x ) d u = f ( x ) + C then f ( x ) =

If ∫ 1 x ( log ⁡ x ) 2 − 2 log ⁡ x + 10 2 d x = 1 54 tan − 1 ⁡ ( f ( x ) ) + 3 ( log ⁡ x − 1 ) g ( x ) + C then f ( e ) + g ( e ) = − − −

∫ x 3 x + 1 d x is equal to

∫ cos ⁡ 2 x − cos ⁡ 2 θ cos ⁡ x − cos ⁡ θ d x is equal to

∫ 1 1 + sin ⁡ x d x =

∫ sin ⁡ 2 x d ( tan ⁡ x ) is equal to

∫ x − 1 x x + 1 d x is equal to

If ∫ cos ⁡ 4 x + 1 cot ⁡ x − tan ⁡ x d x = A cos ⁡ 4 x + B , then

∫ x 3 − 1 x 4 + 1 ( x + 1 ) d x =

If f ‘ ( x ) = 1 − x + x 2 + 1 and f ( 0 ) = − 1 + 2 2 then f ( 1 ) =

∫ dx 2 + sin ⁡ x + cos ⁡ x is equal to

∫ cos ⁡ x − sin ⁡ x 1 + sin ⁡ 2 x dx is equal to

∫ | x | dx is equal to

∫ − 1 4 ( 2 x − 3 ) dx is equal to

∫ 0 π / 2 sin ⁡ x 1 + cos 2 ⁡ x dx is equal to

If f and g are defined as f(x) = f (a -x) and g(x) + g (a -x) = 4, then ∫ 0 a f ( x ) g ( x ) dx is equal to

The value of the integral ∫ 1 / n ( an − 1 ) / n x a − x + x dx is

100 ∫ 0 1 { x } dx where {x } denotes the fraction part of x.

If ∫ d x x 4 − x 2 = 1 x + log ⁡ | f ( x ) | + C then f ( x ) is given by

∫ cos ⁡ 4 x − 1 cot ⁡ x − tan ⁡ x dx is equal to

If ∫ x + sin ⁡ x 1 + cos ⁡ x d x = f ( x ) tan ⁡ x 2 + C then

Let f ( x ) = x 1 + x n 1 / n for n ≥ 2 and g ( x ) = f ∘ f … o f n times . Then ∫ x n − 2 g ( x ) d x equal

If I n = ∫ cot n ⁡ x d x , and I 0 + I 1 + 2 I 2 + … + I 8 + I 9 + I 10 = A u + u 2 2 + … + u 9 9 , where u = cot ⁡ x then

If ∫ f ( x ) x 2 − x + 1 d x = 3 2 log ⁡ x 2 − x + 1 + 1 3 tan − 1 ⁡ 2 x − 1 3 + C then f ( x ) is equal to

If θ 1 and θ 2 be respectively the smallest and the largest values of θ ln 0 , 2 π − π which satisfy the equation 2 cot 2 θ − 5 sin θ + 4 = 0 then ∫ θ 1 θ 2 cos 2 3 θ d θ is equal to:

The value of α for which 4 α ∫ − 1 2 e − α x d x = 5 , is:

The value of I = ∫ 0 2 π x sin 8 x sin 8 x + cos 8 x d x is equal to

I 1 = ∫ 0 4 x – 1 2 d x and I 2 = ∫ 0 π 1 + cos 2 x 2 d x then ( [ . ] denotes the G.I.F.)

If y = y ( x ) and 2 + sin x y + 1 d y d x = – cos x , y ( 0 ) = 1 , then y π 2 equals

T h e v a l u e o f lim n ∞ n 1 ( n + 1 ) ( n + 2 ) + 1 ( n + 2 ) ( n + 4 ) + ⋯ + 1 6 n 2

If I= ∫ − π / 2 π / 2 cos ⁡ x − cos 3 ⁡ x d x t h e n 3 I i s

Evaluate ∫ 0 10 π tan − 1 ⁡ x d x , where [ ⋅ ] represents greatest integer function.

Evaluate ∫ − π 3 π log ⁡ ( sec ⁡ θ − tan ⁡ θ ) d θ .

If f ( 0 ) = 1 , f ( 2 ) = 3 , f ‘ ( 2 ) = 5 , then find the value of ∫ 0 1 x f ′′ ( 2 x ) d x

Evaluate ∫ − π π x sin ⁡ x d x e x + 1

Evaluate ∫ π / 6 π / 3 ( sin ⁡ x ) d x ( sin ⁡ x ) + ( cos ⁡ x )

Evaluate ∫ 0 π x log ⁡ sin ⁡ x d x

If [ x ] denotes the greatest integer less than or equal to x , then the value of the integral ∫ 0 2 x 2 [ x ] d x .

Evaluate ∫ 0 π / 2 x cot ⁡ x d x

If x = ∫ 0 y d t 1 + 9 t 2 and d 2 y d x 2 = a y , then a =

Let f be a real-valued function satisfying f ( x ) + f ( x + 4 ) = f ( x + 2 ) + f ( x + 6 ) then ∫ x x + 8 f ( t ) d t is

∫ − 2 2 sin − 1 sin x + cos − 1 cos x 1 + x 2 1 + x 2 5 d x = ( w h e r e [ . ] r e p r e s e n t s t h e g r e a t e s t i n t e g e r f u n c t i o n )

The value of lim n ∞ 1 n a + 1 n a + 1 + 1 n a + 2 + ⋯ + 1 n b is

∫ s i n 2 x sin 5 x sin 3 x d x =

∫ p x p + 2 q – 1 – q x q – 1 x 2 p + 2 q + 2 x p + q + 1 d x =

∫ e x – 1 d x =

Evaluate ∫ sin 3 ⁡ x d x

Evaluate ∫ sin ⁡ x sin ⁡ 3 x d x

Evaluate ∫ 1 x 2 − x + 1 d x

Evaluate ∫ 1 3 sin ⁡ x + cos ⁡ x d x

Evaluate ∫ x sin ⁡ 3 x d x

Evaluate ∫ x − sin ⁡ x 1 − cos ⁡ x d x

Evaluate ∫ 2 x x 2 + 1 x 2 + 2 d x

∫ x + x 2 + 1 d x equals ( where C is integration Constant) to

If f x = x cos ⁡ x e x 6 sin 5 ⁡ x x 4 sec ⁡ x tan 3 ⁡ x 10 20 , then the value of ∫ − π 2 π 2 f x d x =

If lim n ∞ ∑ k = 0 n n C k n k ( k + 3 ) = e − λ then λ =

Let g ( x ) = ∫ 1 + 2 cos ⁡ x ( cos ⁡ x + 2 ) 2 d x and g ( 0 ) = 0 . Then the value of g ( 3 π / 2 ) is

∫ 0 π / 4 π x − 4 x 2 log 1 + tan x d x =

∫ tan 2 x sec 4 x d x

∫ d x 4 sin 2 ⁡ x + 5 cos 2 ⁡ x = K tan − 1 ⁡ 2 5 tan ⁡ x + C , then K =

If f ( x ) is a polynomial satisfying f ( x ) ⋅ f 1 x = f ( x ) + f 1 x and f ( 3 ) = 82 , then ∫ f ( x ) x 2 + 1 d x =

∫ x 2 1 − x 4 d x

∫ sin 2 x cos 2 x d x sin 5 x + cos 3 x sin 2 x + sin 3 cos 2 x + cos 5 x 2

∫ cos x − 1 sin x + 1 e x d x =

∫ 1 + 2 tan x tan x + sec x 1 / 2 d x

∫ sin 2 x a 2 sin 2 x + b 2 cos 2 x d x

The value of the definite integral ∫ − 1 2 1 2 sin − 1 ⁡ 3 x − 4 x 3 − cos − 1 ⁡ 4 x 3 − 3 x d x =

∫ 0 1 s i n − 1 x 1 − x 2 d x

∫ sin 6 x cos 8 x d x =

∫ cot x cot ( 240 0 − x ) cot ( 240 0 + x ) d x

∫ 1 cos 2 x sin 4 x d x =

∫ e 2 log x 1 + x 6 − 1 d x = − − −

∫ sec x − 1 dx

∫ 1 + 2 x + 3 x 2 + 4 x 3 + ⋯ d x =

∫ x 3 d tan − 1 ⁡ x is equal to

If f ( x ) = ∫ x 2 + sin 2 ⁡ x 1 + x 2 sec 2 ⁡ x d x and f ( 0 ) = 0 , then f ( 1 ) =

If ∫ 2 x 1 − 4 x d x = K sin − 1 ⁡ 2 x + C , then K is equal to

∫ sin 8 ⁡ x − cos 8 ⁡ x 1 − 2 sin 2 ⁡ x cos 2 ⁡ x d x =

∫ 1 e x + e − x 2 d x =

If I = ∫ d x 2 a x + x 2 3 / 2 , then I is equal to

If ∫ d x x 2 x n + 1 ( n − 1 ) / n = − [ f ( x ) ] 1 / n + c , then f ( x ) is

∫ ( x + 1 ) 2 d x x x 2 + 1 is equal to

If ∫ ( x − 1 ) 2 x 4 + x 2 + 1 d x = f ( x ) + C , then the value of lim x ∞ f ( x ) is

∫ sin − 1 ⁡ 2 x 1 + x 2 d x is equal to

If I = ∫ 5 − x 2 + x d x , then I equals

∫ e x 2 tan ⁡ x 1 + tan ⁡ x + cot 2 ⁡ x + π 4 d x is equal to

If ∫ x 2 − 4 x 4 + 9 x 2 + 16 d x = A tan − 1 ⁡ ( f ( x ) ) + B , then the value of f ( 10 ) is

If ∫ 2 cos ⁡ x − sin ⁡ x + λ cos ⁡ x + sin ⁡ x − 2 d x = A ln ⁡ | cos ⁡ x + sin ⁡ x − 2 | + B x + C then the value of λ A B is

Evaluate ∫ x − 1 3 1 + x 1 / 3 1 / 4 dx

∫ x 2 − 1 x 4 + x 2 + 1 dx is equal to

If ∫ f ( x ) dx = f ( x ) + C , then ∫ { f ( x ) } 2 dx is

∫ 1 x − x dx is equal to

∫ e tan − 1 ⁡ x 1 + x 2 dx is equal to

∫ e 2 x − e − 2 x e 2 x + e − 2 x dx is equal to

∫ 2 7 x x + 9 − x dx is equal to

∫ 0 π x 1 + sin ⁡ x dx is equal to

Let I 1 = ∫ a π − a xf ( sin ⁡ x ) dx , I 2 = ∫ a π − a f ( sin ⁡ x ) dx , then I 2 is equal to

If ∫ d x x 5 x 2 − 3 = K tan − 1 ⁡ f ( x ) + C then

The integral ∫ e tan − 1 ⁡ x 1 + x + x 2 1 + x 2 d x is equal to

If the antiderivative of 1 x 2 1 + x 2 is − f ( x ) x + C then f ( x ) is equal to

If it is known that at the point x = 1 two antiderivatives of f ( x ) = e x differ by 2 The difference of these antiderivatives at x = 100 is

if ∫ 3 x + 1 − 7 x − 1 21 x d x = K 1 3 − x + K 2 7 − x + C then

∫ 7 x 13 + 5 x 15 x 7 + x 2 + 1 3 equals

∫ e x 1 − x 1 + x 2 2 d x =

If ∫ x 7 1 + x 4 2 d x = 1 4 log ⁡ 1 + x 4 + f ( x ) + C then

If ∫ e x cos ⁡ 4 x d x = A e 5 x sin ⁡ 4 x + 5 4 cos ⁡ 4 x + C then A is equal to

If f ( x ) = 1 cos 2 ⁡ x 1 + tan ⁡ x then its anti-derivative F ( x ) , F ( 0 ) = 4 is

If f ( x ) = cos ⁡ x then ∫ 2 ( f ( x ) ) 2 − 1 4 ( f ( x ) ) 3 − 3 f ( x ) ) d x is equal to

If ∫ sin ⁡ 2 x cos 4 ⁡ x + sin 4 ⁡ x d x = tan − 1 ⁡ ( f ( x ) ) + C then

For x ∈ ( 0 , 5 π / 2 ) , define f ( x ) = ∫ 0 x t sin ⁡ t d t Then f has

The value of ∫ − 2 2 | 1 − x | d x is

The value of ∫ 0 3 | 2 − x | d x = 5 k t h e n k =

∫ e x ( 1 + sin ⁡ x ) 1 + cos ⁡ x d x is equal to

The value of ∫ 0 1 x 2 ( 1 − x ) 9 d x is

If g ( x ) = ∫ 0 x cos 4 ⁡ t dt then g ( x + π ) equals

The value of ∑ n = 1 1000 ∫ n − 1 n e x − [ x ] d x is ([x] is greatest integer function)

The value of the integral ∫ 0 3 d x x + 1 + 5 x + 1 is

∫ d x a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x ( a , b > 0 ) is equal to

The equation of the tangent to the curve y = ∫ x 2 x 3 d t 1 + t 2 at x = 1 is

Let f be a continuous function on R satisfying f ( x + y ) = f ( x ) + f ( y ) for all x , y ∈ R with f ( 1 ) = 2 and g be a function satisfying f ( x ) + g ( x ) = e x then the value of the integral ∫ 0 1 f ( x ) g ( x ) d x is

The value of ∫ 1 / 2 1 2 x sin ⁡ 1 x − cos ⁡ 1 x d x is

∫ a b ( x − a ) ( b − x ) d x ( b > a ) is equal to

If f ( x ) = e cos ⁡ x sin ⁡ x for | x | ≤ 2 2 otherwise then ∫ − 2 3 f ( x ) d x =

Let I = ∫ 0 π / 2 d x 5 + 3 cos 2 x Statement-1: π 16 ≤ I ≤ π 10 Statement-2: 1 8 ≤ 1 5 + 3 cos 2 x ≤ 1 5

If f ( x ) = ∫ e x 2 ( x − 2 ) ( x − 3 ) ( x − 4 ) d x then f increases on

If f is continuously differentiable function then ∫ 0 2.5 x 2 f ′ ( x ) d x is equal to

If ∫ d x 1 + e x = x + f ( x ) + C , then f ( x ) is equal to

If f ( x ) = x 2 − a 2 then ∫ x 2 f ( x ) d x is

Let f ( x ) be a function satisfying f ′ ( x ) = f ( x ) and f ( 0 ) = 2 . Then ∫ f ( x ) 3 + 4 f ( x ) d x

The value of ∫ cos ⁡ x sin ⁡ x + cos ⁡ x d x is

If ∫ 1 + sec ⁡ x d x = K sin − 1 ⁡ ( f ( x ) ) + C then

Let f be a continuous function [a, b] such that f ( x ) > 0 for all x ∈ [ a , b ] . If F ( x ) = ∫ a x f ( t ) d t then

If f n ( x ) = log ⁡ log ⁡ … log ⁡ x (log is repeated n times) then ∫ x f 1 ( x ) f 2 ( x ) … f n ( x ) − 1 d x is equal to

Let I = ∫ 0 π cos x 1 – sin 2 x d x Statement-1: I = π Statement-2: The integrand can be expressed as X [ 0 , π / 2 ] X [ π / 2 , π ] ; X A being the characteristic function of A.

lim n ∞ 1 n ∑ r = 1 2 n r n 2 + r 2 equals

The value of ∫ 0 1 x 2 ( 1 − x ) 9 d x is

If n ∈ N , the value of ∫ 0 n [ x ] d x (where ( x ) is the greatest integer function) is

If ∫ 0 π / 4 x sin ⁡ x cos 3 ⁡ x d x = π 4 + A , then A is equal to

The value of I = ∫ 0 π x d x 4 cos 2 ⁡ x + 9 sin 2 ⁡ x is

The value of I = 2 ∫ sin ⁡ x sin ⁡ ( x − π / 4 ) d x is

The antiderivative of f ( x ) = log ⁡ ( log ⁡ x ) + ( log ⁡ x ) − 2 whose graph passes through ( e , e ) is

If x 2 f ( x ) + f ( 1 / x ) = 0 for all x ∈ R ~ { 0 } then ∫ cos θ sec θ f ( x ) d x =

The value of ∫ 0 π / 2 d x 1 + tan 3 ⁡ x is

The value of ∫ 1 / e tan ⁡ x t 1 + t 2 d t + ∫ 1 / e cot ⁡ x d t t 1 + t 2 is

If a is a positive integer, then the number of values of a satisfying ∫ 0 π 2 { a 2 cos 3 x 4 + 3 4 cos x + a sin x – 20 cos x } d x ≤ – a 2 3

If f a + b + 1 − x = f x , for all x, where a and b are fixed positive real numbers, then 1 a + b ∫ a b x f x + f x + 1 d x is equal to

If ∫ cos x d x sin 3 x 1 + sin 6 x 2 / 3 = f ( x ) 1 + sin 6 x 1 / λ + c where c is a constant of integration, then λ f π 3 is equal to:

The integral ∫ d x ( x + 4 ) 8 7 ( x – 3 ) 6 7 is equal to: (where C is a constant of integration)

If for all real triplets ( a , b , c ) , f ( x ) = a + b x + c x 2 ; then ∫ 0 1 f ( x ) d x is equal to:

If f ( x ) = ∫ 5 x 4 + 4 x 5 x 5 + x + 1 2 d x and f ( 0 ) = 0 , then f ( 1 ) =

The value of ∫ 0 9 { x } d x , where { x } denotes the fractional part of x , is

∫ x x 9 + 1 x 2 + 1 d x = f ( x ) and f ( 0 ) = 0 find the value of f ( 1 )

∫ e cot x sin 2 x 2 log cos e c x + sin 2 x d x

∫ 1 − x 1 + x d x

If ∫ 2 x + 5 7 – 6 x – x 2 d x = A 7 – 6 x – x 2 + B sin – 1 x + 3 4 + C (where C is a constant of integration), then the ordered pair ( A , B ) is equal to:

Let f : R R be a continuous function such that f ( x ) + f ( x + 1 ) = 2 , for all x ∈ R . If I 1 = ∫ 0 8 f ( x ) d x and I 2 = ∫ – 1 3 f ( x ) d x , then the value of I 1 + 2 I 2 is equal to

∫ d x x 1 5 4 + x 4 5 1 2

If f ( x ) is an integrable function in π 6 , π 3 and I 1 = ∫ π / 6 π / 3 sec 2 θ f ( 2 sin 2 θ ) d θ and I 2 = ∫ π / 6 π / 3 cosec 2 θ f ( 2 sin 2 θ ) d θ , then find I 1 2 I 2

The value of lim n ∞ ∑ K = 1 n K n 2 + K 2

E v a l u a t e lim n ∞ 1 2 + 2 2 + 3 3 + ⋯ + n 2 1 3 + 2 3 + 3 3 + ⋯ + n 3 1 6 + 2 6 + 3 6 + ⋯ + n 6

Evaluate ∫ − 1 2 x 3 − x d x

Evaluate ∫ 0 4 π d x cos 2 ⁡ x 2 + tan 2 ⁡ x

Evaluate ∫ − π / 2 π / 2 log ⁡ a − sin ⁡ θ a + sin ⁡ θ d θ , a > 0 .

Evaluate ∫ 0 100 ( x − [ x ] ) d x (where [ re- presents the greatest integer function).

Evaluate ∫ 0 16 π / 3 | sin ⁡ x | d x

Evaluate lim x ∞ ∫ 0 x e x 2 d x 2 ∫ 0 x e 2 x 2 d x

Let I 1 = ∫ π / 6 π / 3 sin ⁡ x x d x , I 2 = ∫ π / 6 π / 3 sin ⁡ ( sin ⁡ x ) sin ⁡ x d x and I 3 = ∫ π / 6 π / 3 sin ⁡ ( tan ⁡ x ) tan ⁡ x d x T hen arrange I 1 , I 2 and I 3 in the decreasing order of their values.

If I = ∫ 0 π 2 cos ⁡ ( sin ⁡ x ) d x , J = ∫ 0 π 2 sin ⁡ ( cos ⁡ x ) d x and K = ∫ 0 π 2 cos ⁡ x d x , then

let f : R R be non-constant differentiable function and satisfies. f ( x ) = x 2 − ∫ 0 1 ( x + f ( t ) ) 2 d t . Then f ( 4 ) is equal to

If ∫ ln ⁡ e x x + 1 + ln ⁡ x x 2 1 + ( x ln ⁡ x ) ln ⁡ e 2 x x d x = f ( x ) + C Where C is arbitrary Constant , then e e f 2 − 1 is =

Let f ( x ) be an even function such that ∫ 0 ∞ f ( x ) d x = π 2 , then the value of the definite integral ∫ 0 ∞ f x − 1 x d x is equal to

If ∫ c o s 4 x + 1 c o t x – t a n x d x = A c o s 4 x + B , t h e n t h e v a l u e o f A i s

If ∫ tan ⁡ x 1 + tan ⁡ x + tan 2 ⁡ x d x = x − K A tan − 1 ⁡ K tan ⁡ x + 1 A + C , ( C is a constant of integration), then the ordered pair ( K , A ) is equal to

The value of l i m n ∞ 1 n 4 ( ∑ r = 1 2 n ( 3 n r 2 + 2 n 2 r ) ) i s e q u a l t o

The value of ∫ 0 2 π cos 2 n ⁡ x cos 2 n ⁡ x + sin 2 n ⁡ x d x is

∫ l o g ( x + 1 + x 2 ) 1 + x 2 d x

∫ 1 x + x 5 d x = f ( x ) + c , t h e n t h e v a l u e o f ∫ x 4 x + x 5 d x =

∫ x sin x s e c 3 x d x =

∫ 1 sin 3 x sin ( x + α ) d x

∫ x 5 1 + x 3 2 / 3 d x = A 1 + x 3 8 / 3 + B 1 + x 3 5 / 3 + c , then

∫ d x x 2 ( 1 + x 5 ) 4 5 =

∫ e tan − 1 ⁡ x 1 + x + x 2 ⋅ d cot − 1 ⁡ x is equal to

∫ 1 ( x − 1 ) 3 ( x + 2 ) 5 1 / 4 d x is equal to

∫ x 2 − 1 x 4 + 3 x 2 + 1 tan − 1 ⁡ x + 1 x d x =

Evaluate ∫ ( 1 + x ) 3 x d x

If ∫ 2 x + 1 − 5 x − 1 10 x d x = = A 1 5 x log ⁡ 1 5 + B 1 2 x log ⁡ 1 2 + c t h e n

Evaluate ∫ sec ⁡ x sec ⁡ x + tan ⁡ x d x .

E v a l u a t e ∫ 2 2 2 x 2 2 x 2 x d x

Evaluate ∫ e 2 x − 1 2 x d x

Evaluate ∫ d x sin ⁡ x cos 3 ⁡ x

∫ 8 x + 13 4 x + 7 d x = = A ( 4 x + 7 ) 3 / 2 + B ( 4 x + 7 ) 1 / 2 + C t h e n A + B =

Evaluate ∫ sin ⁡ 2 x sin ⁡ 3 x d x

Evaluate ∫ 1 1 + e − x d x

Evaluate ∫ 1 x 2 1 + x 2 d x

Evaluate ∫ tan ⁡ x sin ⁡ x cos ⁡ x d x

Evaluate ∫ 1 3 + sin ⁡ 2 x d x

∫ 1 1 + sin ⁡ x + cos ⁡ x d x

Evaluate ∫ x 2 + 4 x 4 + 16 d x

Evaluate ∫ 1 sin 4 ⁡ x + cos 4 ⁡ x d x

Evaluate ∫ e x 4 − e 2 x d x

Evaluate ∫ 4 x + 1 x 2 + 3 x + 2 d x

Evaluate ∫ 1 + x x d x

Evaluate ∫ 3 x 2 tan ⁡ 1 x − x sec 2 ⁡ 1 x d x

Evaluate ∫ log ⁡ ( log ⁡ x ) + 1 ( log ⁡ x ) 2 d x

Evaluate ∫ e x 1 x − 1 x 2 d x

Evaluate ∫ log ⁡ x ( 1 + log ⁡ x ) 2 d x

Evaluate ∫ e 2 x sin ⁡ 3 x d x

Evaluate ∫ sin ⁡ ( log ⁡ x ) d x

Evaluate ∫ x 2 + 2 x + 5 d x

Evaluate ( x − 5 ) x 2 + x d x = ( x 2 + x ) 3 2 3 – A x + 1 2 x + 1 2 2 − 1 2 2 + B log ⁡ x + 1 2 + x + 1 2 2 − 1 2 2 + C

Evaluate ∫ 1 ( x − 3 ) x + 1 d x

Evaluate ∫ 1 − cos ⁡ x cos ⁡ x ( 1 + cos ⁡ x ) d x

Evaluate ∫ x 2 + 1 ( x − 1 ) 2 ( x + 3 ) d x

E v a l u a t e ∫ x 2 ( x 2 + 1 ) ( x 2 + 4 ) d x

∫ x − 11 1 + x 4 − 1 / 2 d x is equal to

I f I n = ∫ ( ln x ) n d x t h e n I n + n I n – 1 =

∫ x x ln ⁡ ( e x ) d x is equal to

If ∫ 1 − x 7 x 1 + x 7 d x = a ln ⁡ | x | + b ln ⁡ x 7 + 1 + c , then

If ∫ g ( x ) d x = g ( x ) , then ∫ g ( x ) f ( x ) + f ′ ( x ) d x is e q u a l t o

If ∫ 1 x 1 − x 3 d x = a log ⁡ 1 − x 3 − 1 1 − x 3 + 1 + b , then a is equal to

Evaluate ∫ sec 2 ⁡ x cosec 2 ⁡ x d x

Evaluate ∫ cos 3 ⁡ x sin ⁡ x d x

Evaluate ∫ tan ⁡ x a + b tan 2 ⁡ x d x

Evaluate ∫ e x ( 1 + x ) cos 2 ⁡ x e x d x

Evaluate ∫ sin 3 ⁡ x cos 5 ⁡ x d x

Evaluate ∫ d x ( 2 x − 7 ) ( x − 3 ) ( x − 4 )

Evaluate ∫ 1 − tan ⁡ x 1 + tan ⁡ x d x

If ∫ 1 sin ⁡ ( x − a ) sin ⁡ ( x − b ) d x = A log sin ( x – a ) sin ( x – b ) + c

Evaluate ∫ tan ⁡ θ d θ = 1 2 tan − 1 ⁡ tan ⁡ θ − 1 2 tan ⁡ θ + 1 2 2 log ⁡ A B + C

Evaluate ∫ x log ⁡ x d x

I = ∫ e 3 log x ( x 4 + 1 ) – 1 d x

∫ cos 2 x – cos 2 θ cos x – cos θ d x =

∫ x 3 x + 1 d x

∫ sin 2 x d ( tan x )

If ∫ 1 cos 3 ⁡ x 2 sin ⁡ 2 x d x = ( tan ⁡ x ) A + C ( tan ⁡ x ) B + k , where k is constant of integration then A + B + C is

Let f ( x ) = maximum { x + | x | , x − [ x ] } , where [ x ] is the greatest integer less than or equal to x. Then ∫ − 2 2 f ( x ) d x =

∫ 0 2 π 1 1 + tan 4 ⁡ x d x is equal to

If ∫ sin ⁡ π 4 − x 2 + sin ⁡ 2 x d x = A tan − 1 ⁡ f x + B , f 0 = 1 , where A , B are constants, then the range of A . f x is

The value of ∫ 0 100 π ∑ r = 1 10 tan ⁡ r x d x is equalto

∫ 2 x 12 + 5 x 9 d x 1 + x 3 + x 5 3 =

If ∫ log e ⁡ log e ⁡ x x log e ⁡ x d x = f ( x ) and f ( e ) = 0 then f e e = … …

If [ x ] is greatest integer function, then ∫ 0 3 x 2 [ x ] d x =

∫ π 10 π sinx dx =

L t n ∞ 1 m + 2 m + 3 m + ….. + n m n m + 1

Lt n ∞ ⁡ n ! n 1 / n is equal to

L t n ∞ ∑ r = 1 n 1 4 n 2 − r 2

L t n ∞ n + 1 + n + 2 + …. + n + n n n

∫ 0 π 2 / 4 cos x d x

∫ 0 1 d x x + x

If ∫ 0 k cos ⁡ x 1 + sin 2 ⁡ x d x = π 4 then k =

∫ 0 1 x tan − 1 x d x

∫ π / 2 3 π / 2 1 1 + cos x d x

∫ 0 k d x 2 + 8 x 2 = π 16 then k =

∫ 0 π / 2 sin . x sin x + cos x d x =

∫ log ⁡ 2 k d x e x − 1 = π 6 then k =

∫ 0 1 x e x x + 1 2 d x

∫ 3 6 x 9 − x + x d x =

If I n = ∫ 0 π / 4 tan n ⁡ θ d θ for n = 1 , 2 , 3 , … … . Then I n − 1 + I n + 1 =

∫ − π / 2 π / 2 log 2 − sin θ 2 + sin θ d θ

∫ 0 π / 2 cos 10 x d x =

∫ 0 π / 2 5 tan x − 3 cot x tan x + cot x d x =

∫ 0 a a 2 − x 2 5 / 2 d x =

∫ 0 π / 2 sin x . d x − x =

∫ 0 2 x 2 d x =

∫ 0 π / 4 tan 4 x + tan 2 x d x

∫ 0 2 x 2 x d x =

If ∫ x 5 e 4 x 3 d x = 1 48 e 4 x 3 f ( x ) + C then f ( x ) =

∫ d x 2 + x − 3 x 2

∫ 1 − x 1 + x d x

∫ 2 − 3 x − 4 x 2 d x

∫ d x 5 sin x + 12 cos x

∫ cot 3 x cos e c − 8 x d x

∫ d x ( 1 + x ) 2010 = 2 1 α ( 1 + x ) α − 1 β ( 1 + x ) β + C where α , β > 0 then α-β is

If ∫ x 5 m − 1 + 2 x 4 m − 1 x 2 m + x m + 1 3 d x = f ( x ) + C then f ( x ) = 0

If I m , n = ∫ cos m ⁡ x sin ⁡ ( n x ) d x then 7 I 4 , 3 − 4 I 3 , 2 is equal to

Let f ( x ) = ∫ e x ( x − 1 ) ( x − 2 ) d x , then f decreases in the inverval

∫ sin 2 x d T a n x =

Let ∫ d x x 2008 + x = 1 p log ⁡ x q 1 + x r + C where p , q , r ∈ N and need not be distinct then the sum of the digits in p + q + r is

∫ e tan x sin x − sec x d x =

∫ cos e c 2 x − 2005 cos 2005 x d x

∫ e x x 4 + x 2 + 1 x 2 + x + 1 d x =

If d d x ( f ( x ) ) = 4 x 3 − 3 x 4 . Such that f ( 2 ) = 0 then f ( x ) is

∫ log x 1 + log x 2 d x =

∫ x 2 x 2 + 1 x 2 + 4 d x =

∫ log log x + 1 log x 2 d x =

∫ sin 2 x cos 2 x 9 − cos 4 2 x d x =

∫ sin ⁡ 101 x sin 99 ⁡ x d x = sin ⁡ ( 100 x ) ( sin ⁡ x ) 2 μ then λ + μ 100 equal to

I = ∫ sec 2 x cos e c 2 x d x

∫ 1 1 + sin x d x .

∫ e x 1 + x cos 2 x e x d x .

∫ sin 3 x d x cos 4 x + 3 cos 2 x + 1 tan − 1 sec x + cos x

∫ sin 8 ⁡ x − cos 8 ⁡ x 1 − 2 sin 2 ⁡ x cos 2 ⁡ x d x is equal to

The value of ∫ 1 + log ⁡ x x x 2 − 1 dx is

∫ sin 8 ⁡ x − cos 8 ⁡ x 1 − 2 sin 2 ⁡ x cos 2 ⁡ x d x is equal to

∫ d x x x n + 1 is equal to

lim n ∞ 1 m + 2 m + 3 m + …. + n m n m + 1 =

∫ 0 2 2 x − 1 d x

∫ 0 6 x x − 1 x − 2 x − 3 x − 4 x − 5 x − 6 d x

∫ 0 102 tan − 1 ⁡ x d x = Where [ . ] = G I F

∫ 0 5 x d x =

∫ 1 1 − sin x d x

∫ sin x sin x − cos x d x

∫ 7 x 8 + 8 x 7 1 + x + x 8 2 d x =

∫ sin 2 x sin 5 x sin 3 x d x =

∫ 3 e x − 5 e − x 4 e x + 5 e − x d x = a x + b ln ⁡ 4 e x + 5 e − x + c then ( a , b ) = − −

∫ sin x sin ( x − a ) d x =

Let f ( x ) be a polynomial satisfying f ( 0 ) = 2 , f ′ ( 0 ) = 3 , f ′′ ( x ) = f ( x ) then f 2 ( 4 ) =

∫ 1 + 2 x + 3 x 2 + 4 x 3 + − − − − d x where 0 < | x | < 1

∫ 3 − 4 tan x 3 tan x + 4 d x = log f ( x ) + c t h e n f ( x ) = − − −

∫ sin 4 ⁡ x cos 2 ⁡ x d x = f ( x ) + 1 4 g ( x ) − 3 x 2 + c then g ( x ) / f ( x ) = ?

∫ x + 5 x + 1 x + 3 5 d x =

∫ 1 1 + x 2020 d x =

∫ 3 x 2 3 log x 3 + 4 d x =

If ∫ 1 1 + x 2 − 1 ( 1 + x ) 2 x d x = − −

∫ cos e c x tan x d x =

∫ cot 3 x d x =

∫ x 5 d sin − 1 x =

I n = ∫ tan n ⁡ x d x then I n + I n + 2 is equal to

I n = ∫ sin n ⁡ x d x , then 5 I 5 − 4 I 3 isequal to

I n = ∫ cot n ⁡ x d x then I 10 is equal to

If I n = ∫ cos ⁡ n x sin ⁡ x d x , then I 10 − I 8 is equal to

If I n = ∫ x n e α x d x for n ≥ 1 , then α I n + n I n − 1 isequal to

If I n = ∫ cosec n ⁡ x d x then 9 I 10 − 8 I 8 is equal to

∫ sin 5 x cos 4 x d x =

∫ t 4 d t 1 − t 10 i s

∫ t t 4 + t 2 − 5 d t is equal to

∫ e x 2 e 2 x + 3 e x + 2 d x is equal to

∫ d x x ( log ⁡ x ) 2 + 4 log ⁡ x + 1

∫ 1 d x x ( log ⁡ x − 2 ) ( log ⁡ x − 3 ) is equal to

∫ d t a 2 sin 2 ⁡ t + b 2 cos 2 ⁡ t is equal to

Integral of sec 2 ⁡ x d x 3 tan 2 ⁡ x + 5 sec 2 ⁡ x + 1 d x is equal to

∫ d x x 2 + x − 2 is equal to

Anti derivative of d x x 2 + 2 x + 3 is equal to

∫ 2 y 4 y + 36 d y is equal to

Anti derivative of 1 ( x + a ) ( x + b ) d x is equal to

Find ∫ cosect − 1 d t

Primitive of ( cos ⁡ x − sin ⁡ x ) 15 − sin ⁡ 2 x is equal to

∫ x 4 a 6 − x 6 d x is equal to

∫ a x 2 d x a − x + a x is equal to

I = ∫ x 2 + x + 1 d x is equal to

∫ x 2 x 6 + 1 is equal to

If J = ∫ e 2 x + a e x d x then J is equal to

I = ∫ sec 3 ⁡ x d x is equal to x ∈ 0 , π 2

∫ 2 cos ⁡ 2 x + sin ⁡ x d x cos ⁡ x is equal to

∫ e x − e [ x ] − x − 2 ⋅ e 3 x 2 d x ∀ x ∈ ( 0 , 1 ) is equal to [where [ x ] is greatest integer of x

∫ 1 x 2 1 + x 2 d x is equal to

∫ x 1 3 2 + x 1 2 2 d x is equal to

∫ x x 2 + 2 x + 2 x + 1 d x is equal to

∫ 1 + x 4 3 x d x is equal to

∫ x − 11 1 + x 2 1 2 d x is equal to

If I = ∫ 1 + x 2 1 − x 2 d x then I is equal to

If I = ∫ 2 − x 2 ( 1 + x ) 1 − x 2 d x then I is equal to

∫ tan ⁡ x sin ⁡ x cos ⁡ x d x is equal to

∫ sin ⁡ 2 t sin 4 ⁡ t + cos 4 ⁡ t d t is equal to

∫ ( sec ⁡ x ) n tan ⁡ x d x is equal to

Antiderivative of tan ⁡ x 3 cos 2 ⁡ x d x is

If P = ∫ sec 2 ⁡ x ( sec ⁡ x + tan ⁡ x ) 10 d x then P is equal to

If J = ∫ sec 3 ⁡ x d x then J is equal to

∫ sec ⁡ x a tan ⁡ x + b d x is equal to

If J = ∫ 1 tan 2 ⁡ x + sec 2 ⁡ x d x then J is equal to

∫ cot ⁡ e x ⋅ cos ⁡ e c 2 e x ⋅ e x d x is equal to

∫ x 2 x sec 2 ⁡ x + tan ⁡ x ( x tan ⁡ x + 1 ) 2 d x is equal to

S = ∫ [ cot ⁡ ( log ⁡ x ) ] 5 cosec 2 ⁡ ( log ⁡ x ) x d x ( x > 0 ) is equal to

∫ [ cot ⁡ ( log ⁡ x ) ] 2 x d x is equal to

∫ cot tan − 1 x 4 cos e c 2 tan − 1 x d x 1 + x 2 tan − 1 x =

∫ cos ⁡ e c 2 2 x 2 x d x is equal to

∫ cot 4 ⁡ y sin 2 ⁡ y d y is equal to

∫ cos ⁡ e c log ⁡ x x d x is equal to

∫ cos ⁡ e c 2 e x e x d x x is equal to

If f ( x ) = ∫ 7 x + x 7 − 1 7 x log ⁡ 7 + 7 x 6 d x then f ( 1 ) − f ( 0 )

∫ cos x sin x − π 4 d x =

If the value of the integral ∫ 0 1 / 2 x 2 1 – x 2 3 / 2 dx is k 6 +π , then k is equal to:

∫ e 3 log ⁡ x x 4 + 1 − 1 d x =

The primitive of the function x | cos ⁡ x | when π 2 < x < π is given by

∫ cosec 4 ⁡ x d x =

∫ sec ⁡ x d x cos ⁡ 2 x =

∫ sin 8 ⁡ x − cos 8 ⁡ x 1 − 2 sin 2 ⁡ x cos 2 ⁡ x d x =

∫ x ln ⁡ a a x / 2 3 a 5 x / 2 b 3 x + ln ⁡ b b x 2 a 2 x b 4 x d x where a , b ∈ R + is equal to

∫ x 2 − 1 x 2 + 1 1 + x 4 d x is equal to

∫ f ( x ) ⋅ ϕ ′ ( x ) − f ′ ( x ) ⋅ ϕ ( x ) f ( x ) ⋅ ϕ ( x ) { log ⁡ ϕ ( x ) − log ⁡ f ( x ) } d x is =

∫ 1 e x + e − x 2 d x =

∫ ln ⁡ ( tan ⁡ x ) sin ⁡ x ⋅ cos ⁡ x d x , is equal to

∫ x 4 − x 1 / 4 x 5 d x is equal to

The value of the integral ∫ x 2 + x x − 8 + 2 x − 9 1 / 10 d x is

∫ x 5 / 2 1 + x 7 d x is

∫ x 3 d x 1 + x 2 is equal to

∫ d x 5 + 4 cos ⁡ x = a tan − 1 ⁡ b tan ⁡ x 2 + C , then

If I = ∫ ( cot ⁡ x − tan ⁡ x ) d x , then I equals

If I = ∫ sin ⁡ 2 x ( 3 + 4 cos ⁡ x ) 3 d x , then I equals

If ∫ ( x ) 5 ( x ) 7 + x 6 d x = a ln ⁡ x k x k + 1 + c , the value of a and k , respectively, are

If ∫ 3 sin ⁡ x + 2 cos ⁡ x 3 cos ⁡ x + 2 sin ⁡ x d x = a x + b ln ⁡ 2 sin x + 3 cos x + C then

∫ d x ( 1 + x ) x − x 2 is equal to

∫ cos ⁡ x − cos 3 ⁡ x 1 − cos 3 ⁡ x d x is equal to

If ∫ d x ( x + 2 ) x 2 + 1 = a ln ⁡ 1 + x 2 + b tan − 1 ⁡ x + 1 5 ln ⁡ | x + 2 | + C then

∫ x 3 − x 1 + x 6 d x is equal to

∫ x 3 − 1 x 3 + x d x is equal to

∫ e − x ( 1 − tan ⁡ x ) sec ⁡ x d x is equal to

If I = ∫ log ⁡ ( cos ⁡ x ) cos 2 ⁡ x , then I equals

∫ sin 4 ⁡ x cos 2 ⁡ x d x is equal to

∫ e tan ⁡ x ( sin ⁡ x − sec ⁡ x ) d x , is equal to

∫ x e x ( 1 + x ) 2 d x is equal to

∫ f ( x ) g ′′ ( x ) − f ′′ ( x ) g ( x ) d x is equal to

∫ sin ⁡ x ⋅ cos ⁡ x ⋅ cos ⁡ 2 x ⋅ cos ⁡ 4 x ⋅ cos ⁡ 8 x d x is equal to

If ∫ x 2 ⋅ e − 2 x d x = e − 2 x a x 2 + b x + c + d , then

∫ e x + e − x + e x − e − x sin ⁡ x 1 + cos ⁡ x d x =

∫ sec 2 ⁡ x − 7 sin 7 ⁡ x d x =

If ∫ cos ⁡ 4 x + 1 cot ⁡ x − tan ⁡ x d x = A cos ⁡ 4 x + B , then the value of A is

∫ dx 1 + 4 x 2 is equal to

If ∫ cos 4 ⁡ x sin 2 ⁡ x dx = Acot ⁡ x + Bsin ⁡ 2 x + C 2 x + D , then

∫ dx 9 − 25 x 2 is equal to

∫ 2 x + 3 x 5 x dx is equal to

∫ sin 3 ⁡ xcos 3 ⁡ xdx is equal to

∫ 1 − x 1 + x dx is equal to

∫ dx ( x + 3 ) 15 / 16 ( x − 4 ) 17 / 16 is equal to

The integral ∫ dx ( x + 4 ) 8 / 7 ( x − 3 ) 6 / 7 is equal to (where C is a constant of integration)

If f ( x ) = ∫ 5 x 8 + 7 x 6 x 2 + 1 + 2 x 7 2 dx , ( x ≥ 0 ) , and f ( 0 ) = 0 , then the value of f(1) is

∫ sin − 7 / 5 ⁡ xcos − 3 / 5 ⁡ dx is equal to

∫ sin 3 ⁡ xcos 3 ⁡ xdx is equal to

∫ dx 9 x 2 + 6 x + 5 is equal to

∫ dx x 2 + 2 x + 2 is equal to

∫ x 2 + 4 x + 1 dx is equal to

∫ 5 x − 2 1 + 2 x + 3 x 2 dx is equal to

∫ x + 2 x 2 + 2 x + 3 dx is equal to

∫ x 2 + 4 x + 7 x 2 + x + 1 dx is equal to

∫ dx 1 + x 2 1 − x 2 is equal to

∫ dx x ax − x 2 is equal to

Evaluate ∫ x − 2 3 1 + x 2 / 3 − 1 .

∫ dx ( x + 2 ) x 2 + 1 is equal to

∫ 3 x − 1 ( x − 1 ) ( x − 2 ) ( x − 3 ) dx is equal to

∫ dx ( x + 2 ) x 2 + 1 is equal to

∫ 3 x − 1 ( x − 1 ) ( x − 2 ) ( x − 3 ) dx is equal to

∫ cos ⁡ 4 xcos ⁡ 7 xdx is equal to

∫ e x 1 + sin ⁡ x 1 + cos ⁡ x dx is equal to

∫ e 2 x sin ⁡ xdx is equal to

If I n = ∫ tan n ⁡ xdx then ( n − 1 ) I n + I n − 2 is equal to

If the primitive of 1 f ( x ) is equal to log ⁡ { f ( x ) } 2 + C then f(x) is

∫ 1 x − x dx is equal to

∫ dx x ( log ⁡ x ) m is equal to

∫ sin 3 ⁡ ( 2 x + 1 ) dx is equal to

∫ 10 x 9 + 10 x log e ⁡ 10 x 10 + 10 x dx is equal to

∫ ( log ⁡ x − 1 ) 1 + ( log ⁡ x ) 2 2 dx is equal to

∫ cos ⁡ log e ⁡ x dx is equal to

∫ e x ( 1 + x ) cos 2 ⁡ e x x dx is equal to

The value of 2 ∫ sin ⁡ xdx sin ⁡ x − π 4 is equal to

∫ dx e x + e − x + 2 is equal to

∫ x 1 − x 3 dx is equal to

∫ sin − 1 ⁡ x 1 − x 2 3 / 2 dx is equal to

If ∫ log ⁡ x + 1 + x 2 1 + x 2 dx = gof ⁡ ( x ) + constant, then

∫ tan − 1 ⁡ x dx is equal to

∫ x e x dx is equal to

∫ dx a 2 + x 2 3 / 2 is equal to

∫ 3 x 9 x − 1 dx is equal to

∫ dx x ( log ⁡ x ) ( log ⁡ log ⁡ x ) … ( log ⁡ log ⁡ … ⏟ 8 times x ) is equal to

∫ x 2 ( ax + b ) − 2 dx is equal to

∫ cos − 1 ⁡ 1 x dx is equal to

∫ x 2 − 1 x 4 + 3 x 2 + 1 tan − 1 ⁡ x + 1 x dx is equal to

∫ mx m + 2 n − 1 − nx n − 1 x 2 m + 2 n + 2 x m + n + 1 dx is equal to

∫ dx x x n + 1 is equal to

∫ dx cos ⁡ x + 3 sin ⁡ x is equal to

∫ ( 1 − cos ⁡ x ) cosec 2 ⁡ xdx is equal to

∫ e x 1 + e x 2 + e x dx is equal to

∫ cos ⁡ 2 x − cos ⁡ 2 α cos ⁡ x − cos ⁡ α dx is equal to

If ∫ 2 1 + sin ⁡ x dx = − 4 cos ⁡ ( ax + b ) + C , then the value of a, b are respectively

∫ dx sin ⁡ x − cos ⁡ x + 2 is equal to

If ∫ cos ⁡ 4 x + 1 cot ⁡ x − tan ⁡ x dx = Acos ⁡ 4 x + B then

∫ cos ⁡ 2 cot − 1 ⁡ 1 − x 1 + x dx is equal to

∫ ( tan ⁡ x + cot ⁡ x ) dx is equal to

If f ( x ) = cos ⁡ x − cos 2 ⁡ x + cos 3 ⁡ x − … ∞ then ∫ f ( x ) d ¨ x is equal to

∫ 3 x 9 x − 1 dx is equal to

∫ dx 8 + 3 x − x 2 is equal to

∫ xe x ( 1 + x ) 2 dx is equal to

∫ x 2 + 3 x dx is equal to

∫ 3 sin ⁡ x + 2 cos ⁡ x 3 cos ⁡ x + 2 sin ⁡ x dx is equal to

∫ x 2 − 1 x 4 + x 2 + 1 dx is equal to

∫ x − 3 ( x − 1 ) 3 ⋅ e x dx is equal to

If ∫ 2 e x + 3 e − x 3 e x + 4 e − x dx = Ax + Blog ⁡ 3 e 2 x + 4 + C , then

∫ e x sec ⁡ x ( 1 + tan ⁡ x ) dx is equal to

∫ cos ⁡ x − 1 sin ⁡ x + 1 e x dx is equal to

∫ x + 2 x + 4 2 e x dx is equal to

∫ 1 + x − x − 1 e x + x − 1 dx is equal to

If I n = ∫ ( log ⁡ x ) n dx , then I n + nI n − 1 is equal to

The value of ∫ dx x + x 3 is

∫ dx a 2 sin 2 ⁡ x + b 2 cos 2 ⁡ x is equal to

∫ dx cos 3 ⁡ x 2 sin ⁡ 2 x is equal to

if the integral ∫ 5 tan ⁡ x tan ⁡ x − 2 dx = x + alog | sin ⁡ x − 2 cos ⁡ x | + k , then a is equal to

∫ e 6 log ⁡ x − e 5 log ⁡ x e 4 log ⁡ x − e 3 log ⁡ x dx is equal to

∫ xsin − 1 ⁡ xdx is equal to

∫ dx ( x + 2 ) x 2 + 1 is equal to

∫ 3 x − 1 ( x − 1 ) ( x − 2 ) ( x − 3 ) dx is equal to

∫ dx ( x + 2 ) x 2 + 4 x + 8 is equal to

∫ 1 − x 2 x ( 1 − 2 x ) dx is equal to

∫ cos ⁡ 4 xcos ⁡ 7 xdx is equal to

∫ e x 1 + sin ⁡ x 1 + cos ⁡ x dx is equal to

∫ xcos ⁡ x + 1 2 x 3 e sin ⁡ x + x 2 dx

∫ e x 1 1 + x 2 + 1 − 2 x 2 1 + x 2 5 dx

∫ ( 1 + x ) sin ⁡ x x 2 + 2 x cos 2 ⁡ x − ( 1 + x ) sin ⁡ 2 x dx

∫ cosec 2 ⁡ x − 2005 cos 2005 ⁡ x dx

If ∫ ( x ) 5 dx ( x ) 7 + x 6 = λlog ⁡ x a x a + 1 + C then a + λ

If ∫ x 7 1 − x 2 5 dx = 1 λ x 8 1 − x 2 4 + C then λ is equal to

If ∫ cos 3 ⁡ x sin 11 ⁡ x dx = − 2 Atan − 9 / 2 ⁡ x + Btan − 5 / 2 ⁡ x + C then 1 A + 1 B is equal to

If ∫e ax cos ⁡ bx = e 2 x 29 f ( x ) + C and f ′′ ( x ) = kf ( x ) then k is equal to

If ∫ cot ⁡ x sin ⁡ xcos ⁡ x dx = P cot ⁡ x + Q then the value of P 4 is

If f ( x ) = lim n ∞ 2 x + 4 x 3 + … + 2 nx 2 n − 1 ( 0 < x < 1 ) , then ∫ f ( x ) dx is equal to

∫ ( x + 1 ) 3 / 2 sin − 1 ⁡ ( log ⁡ x ) + cos − 1 ⁡ ( log ⁡ x ) dx is equal to

Let x 2 ≠ nπ − 1 , n ∈ N .The value of ∫ x 2 sin ⁡ x 2 + 1 − sin ⁡ 2 x 2 + 1 2 sin ⁡ x 2 + 1 + sin ⁡ 2 x 2 + 1 dx is equal to

Given, f ( x ) = 0 x 2 − sin ⁡ x cos ⁡ x − 2 sin ⁡ x − x 2 0 1 − 2 x 2 − cos ⁡ x 2 x − 1 0 ‘ then ∫ f ( x ) dx is equal to

If an anti-derivative of f(x) is e x and that of g(x) is cos x, then ∫ f ( x ) cos ⁡ xdx + ∫ g ( x ) e x dx is equal to

∫ 1 ( x − 1 ) 3 ( x + 2 ) 5 1 / 4 dx is equal to

∫ | x | log ⁡ | x | dx is equal to ( x ≠ 0 )

Let f(x) be a function such that, f ( 0 ) = f ′ ( 0 ) = 0 , f ′′ ( x ) = sec 4 ⁡ x + 4 then the function is

∫ ( sin ⁡ 2 x − cos ⁡ 2 x ) dx = 1 2 sin ⁡ ( 2 x − a ) + b , then

∫ x x ( 1 + log ⁡ | x | ) dx is equal to

∫ dx x 2 x 4 + 1 3 / 4 = A x 4 + 1 x 4 B + C

If ∫ ( 2 x − 1 ) dx x 4 − 2 x 3 + x + 1 = Atan − 1 ⁡ f ( x ) 3 + C then

∫ − 5 5 | x + 2 | dx is equal to

∫ 0 π / 2 sin ⁡ x sin ⁡ x + cos ⁡ x dx is equal to

∫ − 3 / 2 10 { 2 x } dx where {.} denotes the fractional part of x, is equal to

∫ 0 4 π | sin ⁡ x | dx is equal to

The value of ∫ − π / 2 π / 2 x 3 + xcos ⁡ x + tan 5 ⁡ x + 1 dx is

∫ 1 2 x 3 − 1 dx where [.] denotes the greatest integer function, is equal to

∫ 0 nπ + w | sin ⁡ x | dx , where n ∈ N and 0 ≤ w < π is equal to

∫ 0 1 dx e x + e − x dx is equal to

If ∫ 0 1 e t 1 + t dt = a , then ∫ 0 1 e t ( 1 + t ) 2 dt is equal to

d dx ∫ x 2 x 3 1 log ⁡ t dt is equal to

The value of ∫ − 1 3 tan − 1 ⁡ x x 2 + 1 + tan − 1 ⁡ x 2 + 1 x dx is

∫ − 3 3 dx 1 + e x 1 + x 2 is equal to

If ∫ 0 1 e t dt t + 1 = a then, ∫ b − 1 b e − t dt t − b − 1 is equal to

∫ 0 1 tan − 1 ⁡ 1 x 2 − x + 1 is equal to

∫ 0 π log ⁡ ( 1 + cos ⁡ x ) dx is equal to

∫ 0 1 sin ⁡ 2 tan − 1 ⁡ 1 + x 1 − x dx is equal to

The value of ∫ 0 1 log ⁡ sin ⁡ πx 2 dx is equal to

∫ 0 π / 2 sin 3 / 2 ⁡ x sin 3 / 2 ⁡ x + cos 3 / 2 ⁡ x dx is equal to

∫ 0 4 | x − 1 | dx is equal to

∫ − a a a − x a + x dx is equal to

∫ − 2 3 x 2 − 1 dx is equal to

∫ − 1 1 x 3 + | x | + 1 x 2 + 2 | x | + 1 dx is equal to

∫ − 2 2 | xcos ⁡ πx | dx is equal to

∫ 0 2 π ( sin ⁡ x + | sin ⁡ x | ) dx is equal to

∫ 0 π cos ⁡ 2 x + 1 2 dx is equal to

∫ 0 π / 2 sin ⁡ 2 x log ⁡ tan ⁡ x dx is equal to

∫ − 1 1 log ⁡ x + x 2 + 1 dx is equal to

∫ 0 π e sin 2 ⁡ x cos 3 ⁡ xdx is equal to

If f : R R and g : R R are one to one, real valued functions, then the value of the integral ∫ − π π [ f ( x ) + f ( − x ) ] [ g ( x ) − g ( − x ) ] dx is

∫ − 2 2 | [ x ] | dx is equal to

Suppose f is such that f(-x) = -f(x) for every real x and ∫ 0 1 f ( x ) dx = 5 then ∫ − 1 0 f ( t ) dt is equal to

If P = ∫ 0 3 π f cos 2 ⁡ x dx and Q = ∫ 0 π f cos 2 ⁡ x dx then

lf f(x) is continuous for all real values of x, then ∑ n = 1 10 ∫ 0 1 f ( r − 1 + x ) dx

Value of ∫ 1 5 { x + 2 x − 1 + x − 2 ( x − 1 ) } dx is

The value ,of ∫ 0 2 x 2 dx where [.] is the greatest integer function

lf f(x) is a function satisfying f 1 x + x 2 f ( x ) = 0 for all non-zero x, then ∫ sin ⁡ θ cosec ⁡ θ f ( x ) dx is equal to

The value of the definite integral ∫ a + 2 π a + 5 π / 2 sin − 1 ⁡ ( cos ⁡ x ) + cos − 1 ⁡ ( sin ⁡ x ) dx is equal to

The value of ∫ 0 sin 2 ⁡ x sin − 1 ⁡ t dt + ∫ 0 cos 2 ⁡ x cos − 1 ⁡ t dt is

lim n ∞ 1 n + 1 + 1 n + 2 + … + 1 6 n is equal to

The value of integral ∑ k = 1 n ∫ 0 1 f ( k − 1 + x ) dx is

If ∫ ( 1 − tan ⁡ 3 x ) 2 d x = 1 2 [ tan ⁡ 3 x + log ⁡ f ( x ) ] + C then f ( x ) is given by

∫ x 5 1 + x 3 2 3 d x is equal to

Let P ( x ) be a polynomial of degree n with leading coefficient 1 . Let v ( x ) be any function and v 1 ( x ) = ∫ v ( x ) d x v 2 ( x ) = ∫ v 1 ( x ) d x … v n + 1 ( x ) = ∫ v n ( x ) d x then ∫ P ( x ) v ( x ) d x is equal to

∫ x 3 − 1 4 x 3 − x d x is equal to

The function f whose graph passes through (0 ,0 ) and whose derivative is cos 4 ⁡ x + sin 4 ⁡ x cos 2 ⁡ x − sin 2 ⁡ x is given by

∫ e x 1 − x 1 + x 2 2 d x is

Let f : ( 11 , ∞ ) ( 0 , ∞ ) be given by f ( x ) = ∏ l = 1 10 1 ( x − l ) where for real number a 1 . . . . . . . . . . a n ∏ l = 1 n a l denotes the product a 1 × a 2 ⋯ × a n Statement 1: ∫ f ( x ) d x = ∑ l = 1 10 ( − 1 ) l log ⁡ | x − l | ( l − 1 ) ! ( 10 − l ) ! Statement 2: For x ∈ [ 11 , ∞ ) f ( x ) = ∑ l = 1 10 A l x − l where A 1 = ∏ j = 1 10 j l − j l = 1 , 2 , 10

If f ( x ) = log ⁡ x x 3 , then its antiderivative F ( x ) given by

∫ 2 x + 1 − 5 x − 1 10 x d x equals

The integral of f ( x ) = 3 x cos ⁡ x is given by

If ∫ d x sin ⁡ 2 x − 2 sin ⁡ x = C − 1 4 log ⁡ | f ( x ) | + 1 8 sin 2 ⁡ x 2 then f ( x ) is equal to

If ∫ x + 1 x 2 + x + 1 d x x 2 + x + 1 = k x − 1 x 2 + x + 1 + C then the value of k

The value of ∫ 1 + sin ⁡ x 1 − sin ⁡ x d x is

∫ 3 3 3 x 3 3 x 3 x d x is equal to

∫ d x a 2 − b 2 x 2 3 / 2 equals

The value of ∫ d x 4 x − 3 − x 2 is equal to

If ∫ 1 − 5 sin 2 ⁡ x cos 5 ⁡ x sin 2 ⁡ x d x = f ( x ) cos 5 ⁡ x + C , then f ( x ) :

Let f be a continuous function satisfying f ( x + y ) = f ( x ) + f ( y ) , for each x , y ∈ R and f ( 1 ) = 2 then ∫ f ( x ) tan − 1 ⁡ x 1 + x 2 2 d x is equal to

If I = ∫ d x x 3 1 + x 6 2 / 3 = f ( x ) 1 + x − 6 1 / 3 + C , where C is a constant of integration, then f ( x ) is equal to:

The value of ∫ d x ( x − 1 ) 3 ( x + 2 ) 5 4 is

If f ( x ) = lim n ∞ x n − x − n x n + x − n , x > 1 then ∫ x f ( x ) log ⁡ x + 1 + x 2 1 + x 2 dx is

∫ x 4 x 2 + 1 d x is equal to

The value of ∫ sin ⁡ x 3 d x is

The value of ∫ log ⁡ ( 1 − x ) d x is

The value of ∫ ( x + 1 ) x 1 + x e x d x is

If ∫ ( sin ⁡ 2 x − cos ⁡ 2 x ) d x = 1 2 sin ⁡ ( 2 x + k ) + C then

If ∫ 2 x − sin − 1 ⁡ x 1 − x 2 d x = C − 2 1 − x 2 − 2 3 f ( x ) then f ( x ) is equal to

The function f whose graph passes through ( π / 4 , 0 ) and whose derivative is log ⁡ ( tan ⁡ x ) sin ⁡ x cos ⁡ x is given by

The function f whose graph passes through ( 4 , – 20 ) and whose derivative is cos ⁡ ( 4 − x ) is given by

If ∫ cos ⁡ x cos ⁡ 2 x cos ⁡ 5 x d x = A 1 sin ⁡ 2 x + A 2 sin ⁡ 4 x + A 3 sin ⁡ 6 x + A 4 sin ⁡ 8 x + C then

If ∫ d x 1 + x + 1 3 = 3 2 ( x + 1 ) 2 / 3 − 3 ( x + 1 ) 1 / 3 + f ( x ) + C then f ( x ) is equal to

If ∫ d x cos 6 ⁡ x + sin 6 ⁡ x = tan − 1 ⁡ ( k tan ⁡ 2 x ) + C , then

If ∫ d x 1 + e x = x + Klog ⁡ 1 + e x + C then K is equal to

If ∫ s i n 2 ⁡ x 1 + s i n 2 ⁡ x d x = x − K t a n − 1 ⁡ ( M t a n ⁡ x ) + C then

If ∫ tan 4 ⁡ x d x = K tan 3 ⁡ x + L tan ⁡ x + f ( x ) , then

If f ( x ) = x , g ( x ) = e x − 1 and h ( x ) = tan − 1 ⁡ x then the anti derivative of ( f o g ) ( x ) is

The value of integral ∫ d x cos 3 ⁡ x sin ⁡ 2 x can be expressed as

If f ( x ) = 4 x 2 + 4 x − 3 then ∫ x + 3 f ( x ) d x is equal to

If f ( x ) = 1 + 3 x log ⁡ 3 and F ( x ) be its antiderivative which assume the value 7 for x = 2 . The value of x for which the curve F ( x ) cuts the abscissa axis is

If f ( x ) = cos ⁡ x and g ( x ) = sin ⁡ x then ∫ log ⁡ f ( x ) ( f ( x ) ) 2 d x is

Let f ( x ) = tan ⁡ x sin ⁡ x cos ⁡ x and F ( x ) is its anti-derivative, if F ( π / 4 ) = 6 then F ( x ) is equal to

Let f ( x ) = 1 4 − 3 cos 2 ⁡ x + 5 sin 2 ⁡ x and if its antiderivative F ( x ) = ( 1 / 3 ) tan − 1 ⁡ ( g ( x ) ) + C then g ( x ) is equal to

f ( x ) = ∫ d x sin 4 ⁡ x is a

The value ∫ d x x 4 + 5 x 2 + 4 IS

If ∫ ( x − 1 ) 2 x 2 + 1 2 d x = tan − 1 ⁡ x + g ( x ) + K then g ( x ) is equal to

If the antiderivative of f ( x ) = sin ⁡ x sin 2 ⁡ x + 4 cos 2 ⁡ x is ( 1 / 3 ) tan − 1 ⁡ ( ( 1 / 3 ) g ( x ) ) + C then g ( x ) is equal to

If f ( x ) d x = ϕ ( x ) i.e. ϕ is a function such that ϕ ′ ( x ) = f ( x ) , then ∫ x 9 f x 5 d x is equal to

The value of integral ∫ d x x 1 − x 3 is given equal to

The integral ∫ cos ⁡ 4 x − 1 cot ⁡ x − tan ⁡ x is equal to

∫ d x 2 − 3 x − x 2 = ( fog ) ⁡ ( x ) + C then

If ∫ x x 2 − 4 x + 8 d x = K log ⁡ x 2 − 4 x + 8 + tan − 1 ⁡ x − 2 2 + C then the value of K is

The value of ∫ sin ⁡ x − cos ⁡ x sin ⁡ 2 x − 1 / 2 d x is

If ∫ cos ⁡ 4 x + 1 cot ⁡ x − tan ⁡ x = K cos ⁡ 4 x + C , then

If ∫ x log ⁡ ( x + 1 x ) d x = f ( x ) log ⁡ ( x + 1 ) + g ( x ) x 2 + L x + C then

The function f whose graph passes through ( 0 , 7 / 3 ) and whose derivative is x 1 − x 2 isgiven by

The value of ∫ d x 5 + 4 cos ⁡ x is

The value of ∫ sin ⁡ α 1 + cos ⁡ α d α is

∫ d x sin ⁡ x + cos ⁡ x is equal to

∫ d x x x 4 + 1 is equal to

The primitive of 1 ( x − a ) 3 / 2 ( b − x ) 1 / 2 is

If f ( x ) is the primitive of sin ⁡ x 3 log ⁡ ( 1 + 3 x ) tan − 1 ⁡ x 2 e x 3 − 1 ( x ≠ 0 ) then lim x 0 f ′ ( x ) is

The value of ∫ π / 4 π / 2 cot ⁡ θ cosec 2 ⁡ θ d θ is

If the primitive of f ( x ) = 1 3 sin ⁡ x + sin 3 ⁡ x is equal to 1 6 log ⁡ t − 1 t + 1 + 1 12 log ⁡ 2 + t 2 − t + C then

The value of lim n ∞ 1 n sin ⁡ π n + sin ⁡ 2 π n + ⋯ + sin ⁡ n π n is

∫ 0 π / 2 d x 2 cos ⁡ x + 3 is equal to

The value I = ∫ 0 π / 2 1 1 + cot ⁡ x d x is

∫ − π / 2 π / 2 x 14 sin 11 ⁡ x + sin 2 ⁡ x d x is equal to

The value of ∫ − π / 2 π / 2 x sin ⁡ x e x + 1 d x is equal to

If ∫ 0 2 d x x + 1 + ( x + 1 ) 3 is equal to k π then k is equal to

If the primitive of 1 e x − 1 2 is f ( x ) − log ⁡ | g ( x ) | + C then

If f is continuously differentiable function then ∫ 0 2.5 x 2 f ′ ( x ) d x is equal to

If ∫ d x sin 4 ⁡ x + cos 4 ⁡ x = 1 2 tan − 1 ⁡ f ( x ) + C then

If ∫ e x t f ( t ) d t = sin ⁡ x − x cos ⁡ x − x 2 2 for all x ∈ R ~ { 0 } then the value of f ( π / 6 ) is

The value of ∫ 0 1 x 3 ( 1 − x ) 11 d x is equal to

Let f ( x ) = ∫ 1 x 2 − t 2 d t . Then the real roots of the equation x 2 − f ′ ( x ) = 0 are

The value of ∫ 0 2 x x 2 + 1 d x , where [ x ] is the greatest integer less than or equal to x is

Let I = ∫ 0 π x 2 cos x d x and J = ∫ 0 π x sin x d x Statement-1: I = – 2 π Statement-2: I = 2 J

The ∫ sin 2 ⁡ x cos 6 ⁡ x d x is a

The integral ∫ − 1 / 2 1 / 2 [ x ] + 2 log ⁡ 1 + x 1 − x d x equal ((where [x] is greatest integer function)

If f ( x ) = ∫ 0 x sin 8 ⁡ t d t , then f ( x + π ) equals

Let I = ∫ – ∞ ∞ d x 1 + x 2 Statement-1: I = π Statement-2: The integrand is even and lim b ∞ F ( b ) = π , F ( b ) = ∫ 0 b d x 1 + x 2

Suppose that the graph of y = f ( x ) , contains the points (0, 4) and (2, 7). If f’ is continuous then ∫ 0 2 f ′ ( x ) d x is equal to

The value I = ∫ − 1 1 cos − 1 ⁡ x + x 7 − 3 x 5 + 7 x 3 − x cos 2 ⁡ x d x is

The value of lim x 0 ∫ 0 x sin ⁡ t 2 d t sin ⁡ x 2 is

Let f ( x ) = ∫ 1 x e − t 2 / 2 1 − t 2 d t then

A line tangent to the graph of the function y = f ( x ) at the point x = a forms an π / 3 with y – axis and at x = b an angle π / 4 with x – axis then ∫ a b f ′′ ( x ) d x is

The value of I = ∫ − π / 2 π / 2 cos ⁡ x − cos 3 ⁡ x d x is

Let I = ∫ 1 2 10 x 2 x 3 + 1 2 d x and J = ∫ 2 9 d x x 2 Statement-1: I = 35 27 Statement-2: 3 I = 10 J

The function F ( x ) = ∫ 0 x log ⁡ 1 − t 1 + t d t is

If F ( x ) = ∫ 3 x 2 + d d t cos ⁡ t d t then F ′ π 6 is equal to

If ∫ tan 7 x d x = f ( x ) + C then

∫ − 4 − 5 e ( x + 5 ) 2 d x + 3 ∫ 1 / 3 2 / 3 e 9 x − 2 3 2 d x is

If ∫ d x sin ⁡ x cos ⁡ x = log ⁡ | f ( x ) | + C then

A polynomial P is positive for x > 0 , and the area of the region bounded by P ( x ) , the x-axis, and the vertical lines x = 0 and x = K is K 2 ( K + 3 ) / 3 .The polynomial P ( x ) is

If polynomials P and Q satisfies ∫ ( ( 3 x − 1 ) cos ⁡ x + ( 1 − 2 x ) sin ⁡ x ) d x = P cos ⁡ x + Q sin ⁡ x (ignoring the constant of integration) then

The value of the integral ∫ 0 n π + t ( | cos ⁡ x | + |sin x|) dx is

∫ 0 π / 4 log ⁡ 1 + tan 2 ⁡ θ + 2 tan ⁡ θ d θ =

lim n ∞ 1 n + 1 n + 1 + … + 1 3 n =

∫ − a a log ⁡ x + x 2 + 1 d x =

I = ∫ x 2 − 1 x 3 2 x 4 − 2 x 2 + 1 d x equal to

I = ∫ 0 π [ cot ⁡ x ] d x where [ · ] denotes the greatest integer function, is equal to

Let I = ∫ e x e 4 x + e 2 x + 1 d x , J = e 3 x e 4 x + e 2 x + 1 then the value of I – J equals

The value of ∫ π / 4 π / 3 d x sin x + tan x is

Let I = ∫ 0 1 sin ⁡ x x d x and J = ∫ 0 1 cos ⁡ x x d x Then which one of the following is true

The value of ∫ 0 4 3 2 x + 1 d x is

If f ( x ) = 1 2 n when 1 2 n + 1 < x ≤ 1 2 n , n = 0 , 1 , 2 … then l t n ∞ ∫ 1 / 2 n 1 f ( x ) d x

If m ≠ n , m , n ∈ N then the value of ∫ 0 2 π cos ⁡ m x cos ⁡ n x d x is

The value of the integral ∫ − π / 3 π / 3 x sin ⁡ x cos 2 ⁡ x d x is

The value of lim x ∞ ∫ 0 x e x d x 2 ∫ 0 x e 2 x 2 d x is

If ∫ log ⁡ ( log ⁡ x ) + 1 ( log ⁡ x ) 2 d x = x [ f ( x ) − g ( x ) ] + C , then

If f 3 x − 4 3 x + 4 = x + 2 , then ∫ f ( x ) d x is

The difference between the greatest and least values of the function F ( x ) = ∫ 0 x ( t + 1 ) d t on [ 1 , 3 ] is

If f : R R , g : R R are continuous functions then the value of the integral ∫ − π / 2 π / 2 [ ( f ( x ) + f ( − x ) ) ( g ( x ) – g ( – x ) ) ] d x is

The value of ∫ − 1 1 x | x | d x is

The value of ∫ 0 1 8 log 1 + x 1 1 + x 2 d x is

I = ∫ cos ⁡ 2 tan − 1 ⁡ 1 − x 1 + x d x is equal to

The value of ∫ − 3 3 d d x tan − 1 ⁡ 1 x + x 3 d x is

Let p ( x ) be a function defined on R such that p ‘ ( x ) = p ‘ ( 1 – x ) , for all x ∈ [ 0 , 1 ] , p ( 0 ) = 1 and p ( 1 ) = 41 . Then ∫ 0 1 p ( x ) d x equals

The value of ∫ 0 1 max e x , e 1 − x d x equals

If f ( x ) = ( 1 + tan x ) ( 1 + tan ( π / 4 – x ) ) and g ( x ) is a function with domain R, then ∫ 0 1 x 3 g ∘ f ( x ) d x is

The value of ∫ − π / 2 π / 2 cos ⁡ t sin ⁡ ( 2 t − π / 4 ) d t is

Let · denote the greatest integer function then the value of ∫ 0 1 ⋅ 5 x x 2 d x is

The absolute value of ∫ 10 19 sin ⁡ x 1 + x 8 d x is

The lim n ∞ S n if S n = 1 2 n + 1 4 n 2 − 1 + 1 4 n 2 − 4 + … + 1 3 n 2 + 2 n − 1 is

If the integral ∫ 5 tan ⁡ x tan ⁡ x − 2 d x = x + a log ⁡ | sin ⁡ x − 2 cos ⁡ x | + ∣ K then a is equal to

If f ( x ) = cosec ⁡ ( x + π / 3 ) cosec ⁡ ( x + π / 6 ) then the value of ∫ 0 π / 2 f ( x ) d x is

Let f be thrice differential function and if u = − f ′′ ( θ ) sin ⁡ θ + f ′ ( θ ) cos ⁡ θ and v = f ′′ ( θ ) cos ⁡ θ + f ′ ( θ ) sin ⁡ θ then I = ∫ d u d θ 2 + d v d θ 2 1 / 2 d θ is equal to

The equal to integral ∫ 0 1 log ⁡ ( 1 − x + 1 + x ) d x is

Let f ( x ) = { x } the fractional part of x then ∫ − 1 1 f ( x ) d x is equal to

∫ 0 π / 2 min ( sin x , cos x ) d x equals to

If g ( x ) = ∫ 0 x cos ⁡ 4 t d t ,, then g ( x + π ) equals

Let f : [ − 1 , 2 ] [ 0 , ∞ ] be a continuous function such that f ( x ) = f ( 1 − x ) for all x ∈ [ − 1 , 2 ] . Let R 1 = ∫ − 1 2 x f ( x ) d x ., and R 2 be the area of the region bounded by y = f ( x ) , x = − 1 and x = 2 and the x-axis. Then

If ∫ 0 ∞ d x x 2 + 4 x 2 + 9 = k π then the value of k is

The value of I = ∫ log ⁡ 2 log ⁡ 3 x sin ⁡ x 2 sin ⁡ x 2 + sin ⁡ log ⁡ 6 − x 2 d x is

The value of ∫ − π 3 π log ⁡ ( sec ⁡ θ − tan ⁡ θ ) d θ is

The value of ∫ 0 ∞ x log ⁡ x 1 + x 2 2 d x is

Let f : R R be defined by f ( x ) = ∫ 0 1 x 2 + t 2 2 − t d t Then the curve y = f ( x ) is

For a continuous function f , the value ∫ 0 ∞ f x n + x − n log ⁡ x x + 1 1 + x 2 d x is

The numbers A , B and C such that a function of the form f ( x ) = A x 2 + B x + C satisfies the conditions f ′ ( 1 ) = 8 , f ( 2 ) + f ′′ ( 2 ) = 33 and ∫ 0 1 f ( x ) d x = 7 / 3 are

Let f be a continuous function satisfying ∫ − π t 2 f ( x ) + x 2 d x = π + 4 3 t 3 for all t, then f π 2 / 4 is equal to

The equation of the tangent to the curve y = ∫ x x 2 log ⁡ t d t at x = 2 is

If I = ∫ 0 1 / 2 d x 1 − x 2 n for n ≥ 1 the value of I is

I = ∫ sin 10 ⁡ x − cos 8 ⁡ x sin 2 ⁡ x + sin 8 ⁡ x cos 2 ⁡ x − cos 10 ⁡ x 1 − 2 sin 2 ⁡ x cos 2 ⁡ x d x is equal to

If f ( 0 ) = 2 , f ′ ( x ) = f ( x ) , ϕ ( x ) = x + f ( x ) then ∫ 0 1 f ( x ) ϕ ( x ) d x is

The equation of a curve is given by y = f ( x ) where f ′ ( x ) is a continuous function. The tangent at points ( 1 , f ( 1 ) ) , ( 2 , f ( 2 ) ) and ( 3 , f ( 3 ) ) make angles π 6 , π 3 and π 4 respectively with positive x-axis. Then ∫ 2 3 f ′ ( x ) f ′′ ( x ) d x + ∫ 1 3 f ′′ ( x ) d x is

The integral I = ∫ sin 2 ⁡ x cos 2 ⁡ x sin 5 ⁡ x + cos 3 ⁡ x sin 2 ⁡ x + sin 3 ⁡ x cos 2 ⁡ x + cos 5 ⁡ x 2 is equal to

If ∫ − 1 3 / 2 | x sin ⁡ π x | d x = k / π 2 , then the value of k is

If ∫ 0 ∞ e − x 2 d x = π 2 then Statement-1: ∫ 0 ∞ e − x x d x = π Statement-2: lim x ∞ e − x 2 = 0

Let f ( x ) = x e x ( 1 + x ) 2 , x ≠ − 1 Statement-1: The antiderivative F of f ( x ) satisfying F ( 0 ) = 1 is 1 1 + x e x Statement-2 : f ( x ) is of the form g ( x ) + g ′ ( x ) e x for some g ( x ) .

If ∫ 0 ∞ sin ⁡ x x d x = π 2 then Statement-1: ∫ 0 ∞ sin ⁡ a x cos ⁡ b x x d x = π / 2 ( a > b > 0 ) Statement-2: lim x 0 sin ⁡ a x cos ⁡ b x x = a

The value of ∫ − π / 2 π / 2 log ⁡ 2 − sin ⁡ θ 2 + sin ⁡ θ d θ is

Statement-1: ∫ − π / 3 π / 3 x 10 sin 9 ⁡ x d x = 0 Statement-2: f ( x ) = x 2 n is an even function and g ( x ) = sin 2 m + 1 ⁡ x 1x is an odd function, m and n are integers.

The value of lim x 0 ∫ 0 x cos ⁡ t 2 d t x is

Let f ( x ) = 2 − sin ⁡ x 2 + cos ⁡ x Statement- 1: The antiderivative of f is a periodic function of period π . Statement-2 : The period of the function g ( x ) = log ⁡ ( 2 + cos ⁡ x ) + A tan − 1 ⁡ ( B tan ⁡ x / 2 ) + C , A , B , C being contants is 2 π .

The value of the integral ∫ 0 π / 4 sin ⁡ x + cos ⁡ x 3 + sin ⁡ 2 x d x is

Let f ( x ) = 1 1 − sin 4 ⁡ x and F be an antiderivative of f . Statement-1: F ( x ) = 1 2 tan ⁡ x + 1 2 2 tan − 1 ⁡ ( 2 tan ⁡ x ) + C Statement-2 : F is a one-one function of tan x .

If f ( x ) = ∫ 1 / x 2 x 2 cos ⁡ t d t , then f ′ ( 1 ) is equal to

If I 1 = ∫ x 1 d t 1 + t 2 and I 2 = ∫ 1 1 / x d t 1 + t 2 for x > 0 then

The least value of the function F ( x ) = ∫ 0 x ( 3 sin u + 4 cos u ) d u on the interval ( 5 π / 4 , 4 π / 3 ] is

Suppose that f is an odd function and F ( x ) = ∫ a x f ( t ) d t . Statement-1: F is an even function Statement-2: ∫ − a a f ( t ) d t = 0

Using the fact that 0 ≤ f ( x ) ≤ g ( x ) , c < x < d ⇒ ∫ c d f ( x ) d x ≤ ∫ c d g ( x ) d x we can conclude that ∫ 1 3 3 + x 3 lies in the interval

The solution of the equation ∫ 2 x d x x x 2 − 1 = π 12 is given by

If the value of ∫ − 2 2 | x cos ⁡ π x | d x = k / π then the value of k is

The solution of the equation ∫ 2 x d x x x 2 − 1 = π 12 is given by

lim n ∞ 1 n 1 + n 2 n 2 + 1 2 + n 2 n 2 + 2 2 + ⋯ + n 2 n 2 + ( n − 1 ) 2 is equal to

Let f ( x ) = 1 x 6 + x 4 and F be a antiderivative of f such that F ( 1 ) = π 4 + 2 3 Statement-1 F 1 3 = π 6 Statement-2: F ( x ) = tan − 1 ⁡ x + 1 3 x − 1 x 3

The integral ∫ 0 1 / a log ⁡ ( 1 + a x ) 1 + a 2 x 2 d x ( a > 0 ) is equal to

The value of lim n ∞ 1 1 3 + n 3 + 2 2 2 3 + n 3 + ⋯ + n 2 n 3 + n 3 is

Statement-1: ∫ 0 2 π cos 2 m + 1 ⁡ x d x = 0 ( m > 0 ) Statement-2: ∫ 0 2 a f ( x ) d x = 2 ∫ 0 a f ( x ) d x

The value of ∫ 0 α d x 1 – cos ⁡ α cos ⁡ x ( 0 < α < π / 2 ) is

The value of lim n ∞ 1 n + n ( n + 1 ) 2 + n ( n + 2 ) 2 + ⋯ + n ( 2 n − 1 ) 2 is

Let I = ∫ 10 18 cos ⁡ x 1 + x 4 d x Statement-1: | I | < 0.1 Statement-2: cos ⁡ x 1 + x 4 < 0.1

The value of ∫ 0 π x log ⁡ ( sin ⁡ x ) d x is (given that ∫ 0 π / 2 log ⁡ sin ⁡ x d x = − π 2 log ⁡ 2

The mean value of the function f ( x ) = 2 e x + 1 on the interval [0, 2] is

Let F ( x ) = ∫ e − x 2 x 5 d x Statement-1 : If F ( 0 ) = 0 , then F ( 1 ) = 1 − 5 2 e − 1 Statement-2 : F increases on (0, ∞ )

The value of ∫ 0 π sin ⁡ 2 n x sin ⁡ x d x is

For m > 0 , n > 0 , let I m , n = ∫ 0 1 x m ( log ⁡ x ) n d x , then I 5 , 5 is given by

If f ( x ) = x | x | , then for any real number a and b with a < b , the value of ∫ a b f ( x ) d x equals

∫ − π π ( cos ⁡ p x − sin ⁡ q x ) 2 d x where p and q are in tegers is equal to

If I n = ∫ 0 1 cos − 1 ⁡ x n d x then I 6 − 360 I 2 is given by

Let f , g and h be continuous functions on [ 0 , a ] such that f ( x ) = f ( a − x ) , g ( x ) = − g ( a − x ) and 3 h ( x ) − 4 h ( a − x ) = 5 . Then ∫ 0 a f ( x ) g ( x ) h ( x ) d x is equal to

Let I n = ∫ 0 π / 2 x n cos ⁡ x d x then I 8 + 56 I 6 is equal to

Let f : ( 0 , ∞ ) R and F ( x ) = ∫ 0 x f ( t ) d t . If F x 2 = x 2 ( 1 + x ) then f ( 4 ) equals

Statement 1: ∫ e x sin ⁡ x d x = e x 2 ( sin ⁡ x − cos ⁡ x ) + C Statement 2: If the integrand is the form of e x f ( x ) + f ′ ( x ) then the integrals in e x f ( x )

If I 1 = ∫ 1 / e tan x t 1 + t 2 d t and I 2 = ∫ 1 / e cot x d t t 1 + t 2 then the value of I 1 + I 2 is

If f ( x ) = A sin ⁡ π x 2 + B , f ′ ( 1 / 2 ) = 2 and ∫ 0 1 f ( x ) d x = 2 A π then the constants A and B are respectively

If f ′ ( x ) = g ( x ) for f o r a ≤ x ≤ b then ∫ a b f ( x ) g ( x ) d x is equal to

If a function f : [ 0 , 27 ] R is differentiable then for some 0 < α < β < 3 , ∫ 0 27 f ( x ) d x is equal to

Let I 1 = ∫ a b ∫ a x f ( t ) d t d x and I 2 = ∫ a b ( b − x ) f ( x ) d x then

Statement 1: For − 1 < a < 4 ∫ d x x 2 + 2 ( a − 1 ) x + a + 5 = λ log ⁡ | g ( x ) | + C , where λ and C are constants Statement 2: For − 1 < a < 4 , f ( x ) = 1 x 2 + 2 ( a − 1 ) x + a + 5 is a continuous function.

If ∫ π / 2 x 3 − 2 sin 2 ⁡ z d z + ∫ 0 y cos ⁡ t d t = 0 then d y d x at ( π / 2 , π ) is

Statement 1 : I = ∫ f ( x ) g ′ ( x ) − f ′ ( x ) g ( x ) f ( x ) g ( x ) { log ⁡ g ( x ) − log ⁡ f ( x ) } d x = 1 2 log ⁡ g ( x ) f ( x ) 2 + C Statement 2: ∫ ( ϕ ( x ) ) n ϕ ′ ( x ) d x = ( ϕ ( x ) ) n + 1 n + 1 + C

If f ( x ) = ∫ 0 sin 2 ⁡ x sin − 1 ⁡ t d t and g ( x ) = ∫ 0 cos 2 ⁡ x cos − 1 ⁡ t dt then the value of f ( x ) + g ( x )

The least value of the function F ( x ) = ∫ x 2 log 1 / 3 ⁡ t d t , x ∈ [ 1 / 10 , 4 ] is at x =

Solution of the equation ∫ log ⁡ 2 x d x e x − 1 = π 6 are

The difference between the greatest and he least value of the function F ( x ) = ∫ 0 x ( t + 1 ) d t on [2, 3] is

If ∫ d x x 3 ( x − 1 ) 1 / 2 = x − 1 ( 3 x + 2 ) 4 x 2 + K tan − 1 ⁡ x − 1 + C then the value of K is

The integral ∫ log 5 log 7 x cos x 2 cos log 35 − x 2 + cos x 2 d x is equal to

The value of lim m ∞ ∫ 0 π / 2 sin 2 m ⁡ x d x ∫ 0 π / 2 sin 2 m + 1 ⁡ x d x

Let P ( x ) be a polynomial of least degree whose graph has three points of inflextion (–1, –1), (1,1) and a point with abscissa 0 at which the curve is inclined to the axis of abscissas at an angle of 60 ∘

The value of the integral ∫ 0 π / 4 sin ⁡ x + cos ⁡ x 3 + sin ⁡ 2 x d x is

Let f be a positive function and I 1 = ∫ 1 − k k x f ( x ( 1 − x ) ) d x , I 2 = ∫ 1 − k k f ( x ( 1 − x ) ) d x where 2 k − 1 > 0 then I 1 / I 2 is

If S n = e 1 / n n 2 + 2 n 2 e 2 / n + 3 n 2 e 3 / n + ⋯ + 1 n e then lim n ∞ S n is

The value of ∫ 0 1 lim n ∞ ∑ k = 0 n x k + 2 2 k k ! d x is

The value of ∫ x 5 + x 4 + 4 x 3 + 4 x 2 + 4 x + 4 x 2 + 2 5 is equal to

If the equality ∫ 0 x b t cos ⁡ 4 t − a sin ⁡ 4 t t 2 d t = a sin ⁡ 4 x x − 1 holds for all x such that 0 < x < π / 4 then a and b are given by

The inflection points on the graph of function y = ∫ 0 x ( t − 1 ) ( t − 2 ) 2 d t are

The value of ∫ 0 1 x 2 α − 1 log ⁡ x d x , if α = 2 n − 1 2 is

The value of the integral ∫ 0 π / 2 d x 1 + 1 6 sin 2 ⁡ x is

The integral ∫ 0 1 / 2 e x 2 − x 2 ( 1 − x ) 3 / 2 ( 1 + x ) 1 / 2 d x is equal to

The value of ∫ e − 1 e 2 log ⁡ x x d x is

The value of a in ( − π , 0 ) satisfying sin ⁡ α + ∫ α 2 α cos ⁡ 2 x d x = 0 is

The value of ∫ 0 π / 2 1 + 2 cos ⁡ x ( 2 + cos ⁡ x ) 2 d x is

If ∫ x tan − 1 ⁡ x 1 + x 2 d x = 1 + x 2 f ( x ) + K log ⁡ x + x 2 + 1 + C then

The value of ∫ sin ⁡ x sin ⁡ 4 x d x is

The integral I = ∫ 0 2 x 2 d x ( [ t ] denotes the greatest integer less than or equal to t) is equal to:

If f ( x ) = e cos ⁡ x sin ⁡ x for | x | ≤ 2 2 otherwise , then ∫ − 2 3 f ( x ) d x =

Let g ( x ) = 1 , 0 ≤ x < 1 x 3 , 1 ≤ x < 4 x , 4 ≤ x < 9 then ∫ 0 9 g ( x ) d x is

The greatest value of the function F ( x ) = ∫ 4 x t 2 − 8 t + 16 d t on [ 0 , 5 ] is

The value of ∫ − 1 3 { | x − 2 | + [ x ] } d x , where x denotes the greatest integer less than or equal to x is

If f ( x ) = sin ⁡ x + sin ⁡ 2 x + sin ⁡ 3 x sin ⁡ 2 x sin ⁡ 3 x 3 + 4 sin ⁡ x 3 4 sin ⁡ x 1 + sin ⁡ x sin ⁡ x 1 then the value of ∫ 0 π / 2 f ( x ) d x is

If for a continuous function f , ∫ − a a f ( x ) d x = K ∫ 0 a ( f ( x ) + f ( − x ) ) d x then the value of K is

If ∫ d x ( x + 1 ) 2 x 2 + 3 x + 4 = K log ⁡ f ( x ) + C , then

The value of lim x 0 ∫ − x x f ( t ) d t ∫ 0 2 x f ( t + 4 ) d t is (where f is a continuous function and f ( x ) > 0 for all x )

The value of ∫ d x x + x 3 is

The area bounded by the curve y = f ( x ) = x 4 − 2 x 3 + x 2 + 3 , x-axis and ordinates corresponding to minimum of the function f(x) is

∫ π / 4 3 π / 4 d x 1 + cos ⁡ x is equal to

If J m = ∫ 1 e log m ⁡ x d x then J 8 + 8 J 7 is equal to

If ∫ d x 1 + x 2 1 − x 2 = F ( x ) and F ( 1 ) = 0 then for x > 0 , F ( x ) is equal to

Let f 1 ( x ) = ∫ 0 x f ( t ) d t , f 2 ( x ) = ∫ 0 x f 1 ( t ) d t and f 3 ( x ) = ∫ 0 x f 2 ( t ) d t if f 3 ( x ) = A ∫ 0 x f ( t ) ( x − t ) 2 d t then the value of A is

The value of ∫ − π π cos 2 ⁡ x 1 + a x d x , a > 0 is

The value of ∫ 0 π sin n ⁡ x cos 2 m + 1 ⁡ x d x is

The value of ∫ 0 π 2 / 4 sin ⁡ x d x is

If the variables x and y are related by the equation x = ∫ 0 y d x 1 + 9 x 2 and d 2 y d x 2 is proportional to y then the constant of proportionality is

Let f : ( 0 , ∞ ) R and let F ( x ) = ∫ 0 x f ( t ) d t If F x 2 = x 2 ( 1 + x ) then f ( 4 ) equals

The value of the integral ∫ 0 100 π 1 − cos ⁡ 2 x d x is

Let f ( x ) be a polynomial of degree three such that f ( 0 ) = 1 , f ( 1 ) = 2 and 0 is a critical point of f ( x ) such that f ( x ) does not have a local extremum at 0. Then ∫ f ( x ) x 2 + 1 d x is equal to

Let f ( x ) = ∫ 1 x 2 − t 2 d t Then the real roots of the equation x 2 − f ′ ( x ) = 0

Let f be an odd function then ∫ − 1 1 ( | x | + f ( x ) cos ⁡ x ) d x is equal to

The value of ∫ cos 3 ⁡ x + cos 5 ⁡ x sin 2 ⁡ x + sin 4 ⁡ x d x is

If ∫ log ⁡ x + 1 + x 2 1 + x 2 d x = g ∘ f ( x ) + Const. then

The value of ∫ 0 π / 2 sin ⁡ 2 θ sin ⁡ θ d θ is

Whenever a < b , the value of ∫ a b | x | x d x is

The value of lim x ∞ ∫ 0 x tan − 1 ⁡ x 2 d x x 2 + 1 is

If A ( t ) = ∫ − 1 t e − | × | d x , then lim t ∞ A ( t ) is equal to

If f ( x ) = x + 2 2 x + 3 . then ∫ f ( x ) x 2 1 / 2 d x is equal to 1 2 g 1 + 2 f ( x ) 1 − 2 f ( x ) − 2 3 h 3 f ( x ) + 2 3 f ( x ) − 2 + C where

Given I m = ∫ 1 e ( log ⁡ x ) m d x , If I m K + I m − 2 L = e then values of K and L are

Let f be a periodic continuous function with period T > 0 . If I = ∫ 0 T f ( x ) d x Then the value of I 1 = ∫ 4 4 + 4 T f ( 3 x ) d x is

The value of the integral ∫ − 1 / 3 1 / 3 x 4 1 − x 4 cos − 1 ⁡ 2 x 1 + x 2 d x is

The value of ∫ sec ⁡ x d x sin ⁡ ( 2 x + θ ) + sin ⁡ θ is

For 0 < α < π he value of the definite integral ∫ 0 1 d x / x 2 + 2 x cos ⁡ α + 1 is equal to

Let f ( x ) = ∫ 0 x 6 − u 2 d u . Then the real roots of the equation x 2 − f ′ ( x ) = 0 are

If ∫ e 4 x − 1 e 2 x log ⁡ e 2 x + 1 e 2 x − 1 d x = t 2 2 log ⁡ t − t 2 4 + u 2 2 log ⁡ u − u 2 4 + C then

The numbers P, Q and R for which the function f ( x ) = P e 2 x + Q e x + R x satisfies f ( 0 ) = – 1 f ′ ( log ⁡ 2 ) = 31 and ∫ 0 log ⁡ 4 [ f ( x ) − R x ] d x = 39 / 2 are given by

If y = f ( x ) be an invertible function with inverse g and h ( x ) = x f ( x ) , then ∫ f ( a ) f ( b ) g ( x ) d x + ∫ a b f ( x ) d x is equal to

If I is the greatest of the definite integrals I 1 = ∫ 0 1 e − x cos 2 ⁡ x d x , I 2 = ∫ 0 1 e − x 2 cos 2 ⁡ x d x , I 3 = ∫ 0 1 e − x 2 d x , I 4 = ∫ 0 1 e − x 2 / 2 d x then

The value of ∫ 0 [ x ] 2 t 2 [ t ] d t (where [ ] denotes greatest integer function)

If the primitive of sin − 3 / 2 ⁡ x sin − 1 / 2 ⁡ ( x + θ ) is − 2 cosec ⁡ θ f ( x ) + C then

If l ( m , n ) = ∫ 0 1 t m ( 1 + t ) n d t then the expression for l ( m , n ) in terms of l ( m + 1 , n − 1 ) is

The value of the integral ∫ 0 1 x c − 1 log ⁡ x d x , c > 0 is

I f a n = ∫ 0 π / 2 sin 2 ⁡ n x sin ⁡ x d x then a 2 − a 1 , a 3 − a 2 , a 4 − a 3 are in

If x 2 f ( x ) + f ( 1 / x ) = 0 for all x ∈ R − { 0 } then ∫ cos ⁡ θ sec ⁡ θ f ( x ) d x =

If ∫ sin ⁡ x 1 t 2 f ( t ) d t = 1 − sin ⁡ x , then f ( 1 / 3 ) is equal to

The value of the integral ∫ − 1 1 d d x tan − 1 ⁡ 1 x d x is

The value of ∫ 0 π sin ⁡ ( n + 1 / 2 ) x sin ⁡ ( x / 2 ) d x ( n ∈ N ) is

∫ 0 21 [ x ] 4 d x is equal

If, for t > 0 the definite integral ∫ 0 t 2 x f ( x ) d x = 2 5 t 5 then f 4 25 is equal to

The definite integral ∫ − 2 0 x 3 + 3 x 2 + 3 x + 3 + ( x + 1 ) cos ⁡ ( x + 1 ) d x equals

If f ( x ) = ∫ x 2 x 2 + 4 e − t 2 d t , then the function f ( x ) increases in

The definite integral ∫ 0 1 1 − x 1 + x d x is equal to

The integral I = ∫ − 3 / 2 3 / 2 [ x ] + x 3 + log a ⁡ x + x 2 + 1 d x is equal

The value of the integral ∫ 0 ∞ x log ⁡ x 1 + x 2 2 d x is

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