Mathematics

Evaluate $$\displaystyle \int_\cfrac{\pi}{4}^\cfrac{\pi}{2}(\sqrt{\tan x}+\sqrt{\cot x})dx=$$


ANSWER

$$\dfrac{\pi}{\sqrt{2}}$$


SOLUTION
$$\int_{\cfrac{\pi }{4}}^{\cfrac{\pi }{2}} {\left( {\sqrt {\tan x}  + \sqrt {\cot x} } \right)dx} $$
$$ = \int_{\cfrac{\pi }{4}}^{\cfrac{\pi }{2}} {\left( {\sqrt {\tan x}  + \cfrac{1}{{\sqrt {\tan x} }}} \right)dx} $$
$$ = \int_{\cfrac{\pi }{4}}^{\cfrac{\pi }{2}} {\left( {\cfrac{{1 + \tan x}}{{\sqrt {\tan x} }}} \right)dx} $$
putting $$\tan x = {t^2}$$
         $$ \Rightarrow {\sec ^2}xdx = 2tdt$$
         $$ \Rightarrow \left( {1 + {{\tan }^2}x} \right)dx = 2tdt$$
         $$ \Rightarrow \left( {1 + {t^4}} \right)dx = 2tdt$$
         $$ \Rightarrow dx = \cfrac{{2tdt}}{{1 + {t^4}}}$$
$$ = \int_1^\infty  {\cfrac{{\left( {1 + {t^2}} \right)}}{{t\left( {1 + {t^4}} \right)}} \times 2tdt} $$
$$ = 2\int_1^\infty  {\cfrac{{1 + {t^2}}}{{1 + {t^4}}}} dt$$
$$ = 2\int_1^\infty  {\cfrac{{\left( {1 + \cfrac{1}{{{t^2}}}} \right)}}{{\left( {{t^2} + \cfrac{1}{{{t^2}}}} \right)}}dt} $$
$$ = 2\int_1^\infty  {\cfrac{{\left( {1 + \cfrac{1}{{{t^2}}}} \right)}}{{\left( {{t^2} + \cfrac{1}{{{t^2}}} - 2 + 2} \right)}}dt} $$
$$ = 2\int_1^\infty  {\cfrac{{\left( {1 + \cfrac{1}{{{t^2}}}} \right)}}{{{{\left( {t - \cfrac{1}{t}} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}}dt} $$
putting $$t - \cfrac{1}{t} = y$$
     $$\ \Rightarrow \left( {1 + \cfrac{1}{{{t^2}}}} \right)dt = dy$$
$$ = 2\int_0^\infty  {\cfrac{{dy}}{{{y^2} + {{\left( {\sqrt 2 } \right)}^2}}}} $$
$$ = 2 \times \cfrac{1}{{\sqrt 2 }}\left[ {{{\tan }^{ - 1}}\left( {\cfrac{y}{{\sqrt 2 }}} \right)} \right]_0^\infty $$
$$ = \sqrt 2 \left[ {{{\tan }^{ - 1}}\left( \infty  \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right]$$
$$ = \sqrt 2 \left( {\cfrac{\pi }{2}} \right)$$
$$ = \cfrac{\pi }{{\sqrt 2 }}$$
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Single Correct Medium Published on 17th 09, 2020
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