Question
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$

Answer

Let $\text{I}=\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
Here, $\text{f(x)}=\log\sin\text{x}$ Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{f}'\text{(x)}=\cot\text{x}$
let $\text{e}^{\text{x}}\log\sin\text{x = t}$
Diff. both sides w.r.t x
$\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\times\frac{1}{\sin\text{x}}\times\cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big[\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\cot\text{x}\big]\text{dx = dt}$
$\Rightarrow\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx = dt}$
$\therefore \int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx} =\int\text{dt}$
$=\text{t + C}$
$=\text{e}^{\text{x}}\log\sin\text{x}+\text{C}$

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