Mathematics

# Solve $\displaystyle \int {\tan x\,\ln \left( {\cos x} \right)} \,dx$

$-\dfrac {(\ln(\cos x))^2}{2}+C$

##### SOLUTION
consider, $I=\displaystyle \int { \tan { x } } \;\ln \left( \cos { x } \right) dx$

Let $\ln \cos x=t$

On differentiating, we have
$\Rightarrow \dfrac{-\sin x}{\cos x}=dt$

$\Rightarrow -\tan x dx=dt$

$\Rightarrow \displaystyle -\int t.dt=-\dfrac{t^2}{2}+c$

$=-\dfrac{1}{2}\left( \ln \cos x\right)^2+c$

$=-\dfrac { { \left[ \ln \left( \cos { x } \right) \right] }^{ 2 } }{ 2 } +c$

Hence, the answer is $-\dfrac { { \left[ \ln \left( \cos { x } \right) \right] }^{ 2 } }{ 2 } +c.$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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