JEE MAIN - Mathematics (2013 (Offline) - No. 6)

Statement-1 : The value of the integral
$$\int\limits_{\pi /6}^{\pi /3} {{{dx} \over {1 + \sqrt {\tan \,x} }}} $$ is equal to $$\pi /6$$

Statement-2 : $$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx.$$

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Statement- 1 is true; Statement-2 is False.
Statement-1 is false; Statement-2 is true.

Explanation

Let $$I = \int\limits_{\pi /6}^{\pi /3} {{{dx} \over {1 + \sqrt {\tan \,x} }}} $$

$$ = \int\limits_{\pi /6}^{\pi /3} {{{dx} \over {\sqrt {\tan \left( {{\pi \over 2} - x} \right)} }}} $$

$$ = \int\limits_{\pi /6}^{\pi /3} {{{\sqrt {\tan \,x} \,dx} \over {1 + \sqrt {\tan \,x} }}} \,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Also, Given,

$$I$$ $$ = \int\limits_{\pi /6}^{\pi /3} {{{\sqrt {\tan \,x} \,dx} \over {1 + \sqrt {\tan \,x} }}} \,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

By adding $$(1)$$ and $$(2),$$ we get

$$2I = \int\limits_{\pi /6}^{\pi /3} {dx} $$

$$ \Rightarrow I = {1 \over 2}\left[ {{\pi \over 3} - {\pi \over 6}} \right]$$

$$ = {\pi \over {12}},$$ statements -$$1$$ is false

$$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} } $$

It is fundamental property.

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