∫ { 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
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
∫ 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 =
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
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
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
if 2 < x < 3 ∫ ( | x − 1 | | x − 2 | + | x − 3 | ) d x
∫ cos x − c o x 2 x 1 − cos x d x =
∫ tan ( x − α ) tan ( x + α ) tan 2 x d x =
∫ log ( log ( log x ) ) x log x log ( log x ) d u = f ( x ) + C then f ( x ) =
∫ cos 2 x − cos 2 θ cos x − cos θ d x is equal to
∫ 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
∫ − 1 4 ( 2 x − 3 ) dx is equal to
∫ 0 π / 2 sin x 1 + cos 2 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
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
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 =
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 =
Evaluate ∫ sin x sin 3 x d x
Evaluate ∫ 1 3 sin x + 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 lim n ∞ ∑ k = 0 n n C k n k ( k + 3 ) = e − λ then λ =
∫ 0 π / 4 π x − 4 x 2 log 1 + tan x d x =
∫ d x 4 sin 2 x + 5 cos 2 x = K tan − 1 2 5 tan x + C , then K =
∫ 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 =
∫ cot x cot ( 240 0 − x ) cot ( 240 0 + x ) d x
∫ e 2 log x 1 + x 6 − 1 d x = − − −
∫ 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 =
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
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
∫ 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 ∫ 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
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
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 =
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
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
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
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
The integral ∫ d x ( x + 4 ) 8 7 ( x – 3 ) 6 7 is equal to: (where C is a constant of integration)
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
The value of lim n ∞ ∑ K = 1 n K n 2 + K 2
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
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
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 5 1 + x 3 2 / 3 d x = A 1 + x 3 8 / 3 + B 1 + x 3 5 / 3 + c , then
∫ 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 =
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 ∫ 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 ∫ tan x sin x cos x d x
Evaluate ∫ 1 3 + sin 2 x d x
Evaluate ∫ x 2 + 4 x 4 + 16 d x
Evaluate ∫ 1 sin 4 x + cos 4 x d x
Evaluate ∫ 4 x + 1 x 2 + 3 x + 2 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 ∫ 1 ( x − 3 ) x + 1 d x
Evaluate ∫ 1 ( x − 3 ) x + 1 d x
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
I = ∫ e 3 log x ( x 4 + 1 ) – 1 d x
∫ cos 2 x – cos 2 θ cos x – cos θ d x =
∫ 0 2 π 1 1 + tan 4 x d x is equal to
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 =
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
If ∫ 0 k cos x 1 + sin 2 x d x = π 4 then k =
∫ 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 =
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 5 tan x − 3 cot x tan x + cot x d x =
∫ 0 π / 4 tan 4 x + tan 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 ( 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
∫ 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
∫ 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
∫ 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
lim n ∞ 1 m + 2 m + 3 m + …. + n m n m + 1 =
∫ 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
∫ 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 ) = − −
∫ 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 ) = ?
If ∫ 1 1 + x 2 − 1 ( 1 + x ) 2 x d 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
∫ 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
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
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
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
∫ 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
∫ 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 )
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
∫ 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
∫ 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
∫ 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
If ∫ cos 4 x sin 2 x dx = Acot x + Bsin 2 x + C 2 x + D , then
∫ 2 x + 3 x 5 x dx is equal to
∫ sin 3 xcos 3 xdx 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)
∫ 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
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
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
∫ 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
∫ 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
∫ dx a 2 + x 2 3 / 2 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 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
∫ dx 8 + 3 x − x 2 is equal to
∫ xe x ( 1 + x ) 2 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
∫ 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
∫ e 6 log x − e 5 log x e 4 log x − e 3 log x 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
∫ 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 ∫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
∫ 1 ( x − 1 ) 3 ( x + 2 ) 5 1 / 4 dx is equal to
∫ | x | log | x | dx is equal to ( x ≠ 0 )
∫ ( 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
∫ − 2 2 | [ x ] | dx 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
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
∫ x 3 − 1 4 x 3 − x d x is equal to
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 ∫ 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 ) :
The value of ∫ d x ( x − 1 ) 3 ( x + 2 ) 5 4 is
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
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
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 ) = cos x and g ( x ) = sin x then ∫ log f ( x ) ( f ( x ) ) 2 d x is
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
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
The primitive of 1 ( x − a ) 3 / 2 ( b − x ) 1 / 2 is
The value of ∫ π / 4 π / 2 cot θ cosec 2 θ d θ is
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
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
If f ( x ) = ∫ 0 x sin 8 t d t , then f ( x + π ) equals
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
The value of I = ∫ − π / 2 π / 2 cos x − cos 3 x d x is
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
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 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
The value of ∫ 0 1 max e x , e 1 − x d x equals
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 f ( x ) = cosec ( x + π / 3 ) cosec ( x + π / 6 ) then the value of ∫ 0 π / 2 f ( x ) d x is
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
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 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
If f ( 0 ) = 2 , f ′ ( x ) = f ( x ) , ϕ ( x ) = x + f ( x ) then ∫ 0 1 f ( x ) ϕ ( x ) d x is
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
The value of ∫ − π / 2 π / 2 log 2 − sin θ 2 + sin θ d θ is
The value of lim x 0 ∫ 0 x cos t 2 d t x is
The value of the integral ∫ 0 π / 4 sin x + cos x 3 + sin 2 x d x is
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 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
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
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 mean value of the function f ( x ) = 2 e x + 1 on the interval [0, 2] is
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
∫ − 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 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
If f ′ ( x ) = g ( x ) for f o r a ≤ x ≤ b then ∫ a b f ( x ) g ( 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
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
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 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
The value of the integral ∫ 0 π / 4 sin x + cos x 3 + sin 2 x d x 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
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
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
If ∫ d x ( x + 1 ) 2 x 2 + 3 x + 4 = K log f ( x ) + C , then
∫ 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
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
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 ) = ∫ 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
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
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
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
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
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