Statement -1: The value of the integral int_{ | vedclass

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(D) Let $I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{	an x}}$.Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we have:$I =

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Statement $-1$ is false,Statement $-2$ is true.

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(D) Let $I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{	an x}}$.Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,we have:$I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{	an(\frac{\pi}{6} + \frac{\pi}{3} - x)}} = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{	an(\frac{\pi}{2} - x)}} = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\cot x}}$.$I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \frac{1}{\sqrt{	an x}}} = \int_{\pi/6}^{\pi/3} \frac{\sqrt{	an x}}{\sqrt{	an x} + 1} dx$.Adding the two expressions for $I$:$2I = \int_{\pi/6}^{\pi/3} \left( \frac{1}{1 + \sqrt{	an x}} + \frac{\sqrt{	an x}}{1 + \sqrt{	an x}} ight) dx = \int_{\pi/6}^{\pi/3} 1 dx = [x]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.Thus,$I = \frac{\pi}{12}$.Statement $-1$ claims the value is $\frac{\pi}{6}$,which is false. Statement $-2$ is a standard property of definite integrals,which is true.

Statement $-1$: The value of the integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}} = \frac{\pi}{6}$.
Statement $-2$: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$.

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Statement $-1$: The value of the integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}} = \frac{\pi}{6}$.Statement $-2$: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$.