Question

$\purple{ \boxed{ \boxed{\red { \displaystyle \: \tt Prove \: \: That:\int_{0}^{\frac{\pi}{2}} \sin 2 x \log \tan x \: d x=0}}}}$
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Answer 1

Answer:

Hey mate here is your Answer

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Answer 2

Question:-

Prove that:

$\sf \large \int ^{ \frac{\pi}{2} } _{0} \: sin2x \: log(tan \: x)dx = 0.$

Given:-

Integration is

$\sf \large \int ^{ \frac{\pi}{2} } _{0} \: sin2x \: log(tan \: x)dx.$

To Prove:-

$\sf \large \int ^{ \frac{\pi}{2} } _{0} \: sin2x \: log(tan \: x)dx = 0.$

Solution:-

$\sf \large Let, I = \int ^{ \frac{\pi}{2} } _{0} \: sin2x \: log(tan \: x)dx \: \: ...(1)$

$\sf \large = \int^{ \frac{\pi}{2} } _{0} \: sin2( \frac{\pi}{2} - x)log(tan( \frac{\pi}{2} - x)dx$

$\sf \large = \int ^{ \frac{\pi}{2} } _{0} \: sin(\pi - 2x)log(tan( \frac{\pi}{2} - x))dx$

$\sf \large = \int ^{ \frac{\pi}{2} } _{0} \: sin(\pi - 2x)log(cot \: x)dx$

$\sf \large = \int^{ \frac{\pi}{2} } _{0} \: sin2x \: log(cot \: x)dx \: ...(2)$

Adding Equation (1) and (2), we get:

$\sf \large2I = \int ^{ \frac{\pi}{2} } _{0} \: sin2x \: log(tan \: x)dx + \int ^{ \frac{\pi}{2} } _{0} \: sin2x \: log(cot \: x)dx$

$\sf \large = \int ^{ \frac{\pi}{2} } _{0} \: sin2x \: [log(tan \: x)+ log(cot \: x)]dx$

$\sf \large = \int ^{ \frac{\pi}{2} } _{0} \: sin2x[log(tan \: x.cot \: x)]dx$

$\sf \large = \int ^{ \frac{\pi}{2} } _{0} \: sin2x(log1)dx$

$\sf \large = 0$

$\sf \large I = 0.$

Answer:-

${ \boxed{ \sf \large \green{Hence, Proved \: that \sf \large \int ^{ \frac{\pi}{2} } _{0} \: sin2x \: log(tan \: x)dx = 0.}}}$

Hope you have satisfied. ⚘