Mathematics

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


ANSWER

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


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
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