Statement-I : The value of the integral ∫π/6π/3dx1+tanx is equal to π/6Statement-II : ∫abf(x)dx=∫abf(a+b−x)dx

I=∫π/6π/3dx1+tanx=∫π/6π/3dx1+cotx Adding 2I=∫π/6π/3(11+tanx+tanx1+tanx)dx =∫π/6π/31⋅dx=(π3−π6)=π6 ∴I=π12Again Statement-II is true.

Statement-I : The value of the integral ∫π/6π/3dx1+tanx is equal to π/6Statement-II : ∫abf(x)dx=∫abf(a+b−x)dx

I=∫π/6π/3dx1+tanx=∫π/6π/3dx1+cotx Adding 2I=∫π/6π/3(11+tanx+tanx1+tanx)dx =∫π/6π/31⋅dx=(π3−π6)=π6 ∴I=π12Again Statement-II is true.

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Statement-I : The value of the integral $\int_{π / 6}^{π / 3} \frac{d x}{1 + \sqrt{tan x}}$ is equal to $π / 6$
Statement-II : $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$

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Statement-I is true, Statement-II is true, Statement-II is not a correct explanation for Statement-I.

Statement-1 is true, Statement-II is false.

Statement-I is false, Statement-II is true.

Statement-I is true, Statement-II is true, Statement-II is a correct explanation for Statement-I.

answer is C.

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Detailed Solution

$I = \int_{π / 6}^{π / 3} \frac{d x}{1 + \sqrt{tan x}} = \int_{π / 6}^{π / 3} \frac{d x}{1 + \sqrt{cot x}} A d d i n g 2 I = \int_{π / 6}^{π / 3} (\frac{1}{1 + \sqrt{tan x}} + \frac{\sqrt{tan x}}{1 + \sqrt{tan x}}) d x = \int_{π / 6}^{π / 3} 1 \cdot d x = (\frac{π}{3} - \frac{π}{6}) = \frac{π}{6} ∴ I = \frac{π}{12}$