Question:
Write a value of $\int \frac{1+\cot \mathrm{x}}{\mathrm{x}+\log \sin \mathrm{x}} \mathrm{dx}$.
Solution:
let $x+\log \sin x=t$
Differentiating it on both sides we get,
$(1+\cot x) d x=d t-i$
Given that $\int \frac{1+\cot x}{x+\log \sin x} d x$
Substituting $i$ in above equation we get,
$=\int \frac{d t}{t}$
$=\log t+c$
$=\log (x+\log \sin x)+c$
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