Statement-1: The value of the integral _/6^/3 dx1+ x is equal to /6 Statement-2: _a^b f(x)…

Question

Statement-1: The value of the integral $\displaystyle\int_{\pi/6}^{\pi/3} \dfrac{dx}{1+\sqrt{	an x}}$ is equal to $\pi/6$

Statement-2: $\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$

Accepted Answer

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.

Solutions · 1