Question

Evaluate: $\int \cot x\log \sin xdx$

• A. ${\left(\log \sin x\right)}^2$
• B. $\frac{1}{2}\left(\log \sin x\right)$
• C. $\frac{1}{2}{\left(\log co\sec x\right)}^2$
• D. $\frac{1}{2}{\left(\log \sin x\right)}^2$

Answer 1

The correct option is D $\frac{1}{2}{\left(\log \sin x\right)}^2$

Let $I=\int \cot x\log \sin xdx$
Put $\log \sin x=t\Rightarrow \frac{1}{\sin x}\cos xdx=dt\Rightarrow \cot xdx=dt$
Therefore
$I=\int tdt=\frac{1}{2}{t}^2=\frac{1}{2}{\left(\log \sin x\right)}^2$
Hence, option 'D' is correct.