# Evaluate : $\int \large\frac{1+cot x}{x+\log \sin x}$$dx ## 1 Answer Toolbox: • \large\frac{d}{dx}$$(\log \sin x)=\large\frac{1}{\sin x}$
• If $f(x)$ is substituted by $f(t),$ then $f'(x) dx=f'(t)dx$
• Hence $\int f(x) dx=\int f(t)dt$
Step 1:
$I=\int\large\frac{1+\cot x}{x+\log\sin x}$$dx Let x+\log\sin x=t On differentiating w.r.t x we get 1+\large\frac{1}{\sin x}$$\cos xdx=dt$
$(1+\cot x)dx=dt$
Step 2:
Now substituting for t and dt we get,
$I=\int\large\frac{dt}{t}$
On integrating we get,
$\;\;=\log \mid t\mid+c$
$\;\;=\log(x+\log\sin x)+c$