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Question

(C) Let $I = \int \csc^4 x \, dx$. We can write $\csc^4 x$ as $\csc^2 x \cdot \csc^2 x$. Using the identity $\csc^2 x = 1 + \cot^2 x$,we get: $I = \in

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \csc^4 x \, dx$. We can write $\csc^4 x$ as $\csc^2 x \cdot \csc^2 x$. Using the identity $\csc^2 x = 1 + \cot^2 x$,we get: $I = \int \csc^2 x (1 + \cot^2 x) \, dx$ $I = \int \csc^2 x \, dx + \int \cot^2 x \csc^2 x \, dx$ For the second integral,let $u = \cot x$,then $du = -\csc^2 x \, dx$,so $\csc^2 x \, dx = -du$. $I = -\cot x + \int u^2 (-du)$ $I = -\cot x - \frac{u^3}{3} + c$ $I = -\cot x - \frac{\cot^3 x}{3} + c$.

$\int \csc^4 x \, dx = $

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