Question

$\int \frac{\cot x}{\log \left(\sin x\right)}dx=$

• A. $\log \left(\log \right(\sin x\left)\right)+c$
• B. $\log \left(\log \right(\sec x\left)\right)+c$
• C. $\log \left(\log \right(\cot x\left)\right)+c$
• D. $\log \left(\log \right(\cos ecx\left)\right)+c$

Answer 1

The correct option is A $\log \left(\log \right(\sin x\left)\right)+c$

Let $I=\int \frac{\cot x}{\log \left(\sin x\right)}dx$

wkt $\int \cot xdx=\log \left(\sin x\right)+c$

So, $\cot xdx=d\left(\log \left(\sin x\right)\right)$

From def. of integral.

$\int \frac{\cot x.dx}{\log \left(\sin x\right)}$

$=\int \frac{d\left(\log \left(\sin x\right)\right)}{\log \left(\sin x\right)}$

Let $\log \left(\sin x\right)=t$

$\int \frac{dt}{t}=\log \left(t\right)+c$

$t=\log \left(\sin x\right)$

$\Rightarrow I=\log \left(\log \left(\sin x\right)\right)+c$

$\therefore \int \frac{\cot x}{\log \sin x}dx$ $=\log \left(\log \left(\sin x\right)\right)+c$