int cot^4 x , dx is equal to | vedclass

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(A) Step $1$: Use the identity $\cot^2 x = \csc^2 x - 1$. $\int \cot^4 x \, dx = \int \cot^2 x (\csc^2 x - 1) \, dx$ Step $2$: Expand the integral. $=

Vedclass · Accepted Answer

$-\frac{\cot^3 x}{3} + \cot x + x + c$

Vedclass · Answer

(A) Step $1$: Use the identity $\cot^2 x = \csc^2 x - 1$. $\int \cot^4 x \, dx = \int \cot^2 x (\csc^2 x - 1) \, dx$ Step $2$: Expand the integral. $= \int \cot^2 x \csc^2 x \, dx - \int \cot^2 x \, dx$ Step $3$: Substitute $\cot^2 x = \csc^2 x - 1$ in the second integral. $= \int \cot^2 x \csc^2 x \, dx - \int (\csc^2 x - 1) \, dx$ Step $4$: Integrate. For the first part, let $u = \cot x$, then $du = -\csc^2 x \, dx$. $= -\frac{\cot^3 x}{3} - (-\cot x - x) + c$ $= -\frac{\cot^3 x}{3} + \cot x + x + c$

$\int \cot^4 x \, dx$ is equal to

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