Write a value


Write a value of $\int \frac{1+\cot \mathrm{x}}{\mathrm{x}+\log \sin \mathrm{x}} \mathrm{dx}$.


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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