int frac{csc^2 x}{1 + cot x} dx = | vedclass

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Question

(B) Let $I = \int \frac{\csc^2 x}{1 + \cot x} dx$. Substitute $1 + \cot x = t$. Differentiating both sides with respect to $x$,we get $-\csc^2 x dx =

Vedclass · Accepted Answer

$-\log(1 + \cot x) + c$

Vedclass · Answer

(B) Let $I = \int \frac{\csc^2 x}{1 + \cot x} dx$. Substitute $1 + \cot x = t$. Differentiating both sides with respect to $x$,we get $-\csc^2 x dx = dt$,which implies $\csc^2 x dx = -dt$. Substituting these into the integral,we get $I = \int \frac{-dt}{t}$. Integrating,we get $I = -\log|t| + c$. Replacing $t$ with $1 + \cot x$,we get $I = -\log|1 + \cot x| + c$.

$\int \frac{\csc^2 x}{1 + \cot x} dx = $

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