Mathematics

Evaluate: $$\displaystyle \int \frac{\ln (\tan x)}{\sin x\cos x}dx$$ 


ANSWER

$$\displaystyle \frac{1}{2}(\ln (\tan x))^{2}+c$$


SOLUTION
$$\displaystyle I=\int \frac{\ln (\tan x)}{\sin x\:\cos x}dx$$

Let $$\displaystyle t=\ln (\tan x)\Rightarrow dt=\dfrac{\sec ^{2}x}{\tan x}dx=\dfrac{dx}{\sin x\:\cos x}$$

$$\therefore \displaystyle I=\int tdt=\frac{1}{2}t^{2}+C$$

$$\displaystyle I=\frac{1}{2}(\ln (\tan x))^{2}+C$$
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Single Correct Medium Published on 17th 09, 2020
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