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Evaluate : $ \int \large\frac{1+cot x}{x+\log \sin x}$$ dx $

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  • $\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,
On integrating we get,
$\;\;=\log \mid t\mid+c$
$\;\;=\log(x+\log\sin x)+c$
answered Nov 7, 2013 by sreemathi.v
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