Question

$\int \frac{\left(1+\cot x\right)}{\left(x+\log \sin x\right)}dx=?$

Answer 1

Consider the given integral.

$I=\int \left(\frac{1+\cot x}{x+\log \left(\sin x\right)}\right)dx$

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

$\frac{dt}{dx}=1+\frac{1}{\sin x}\times \cos x$

$dt=(1+\cot x)dx$

Therefore,

$I=\int \frac{dt}{t}$

$I=\ln \left(t\right)+c$

On putting the value of $t$, we get

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

Hence, the value of integral is $I=\ln (x+\log \left(\sin x\right))+c$.