int frac{cos^3 x}{sin^2 x + sin x} dx = | vedclass

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Question

(A) Let $I = \int \frac{\cos^3 x}{\sin^2 x + \sin x} dx$.Substitute $\cos^2 x = 1 - \sin^2 x$:$I = \int \frac{(1 - \sin^2 x) \cos x}{\sin^2 x + \sin x

Vedclass · Accepted Answer

$\log|\sin x| - \sin x + c$

Vedclass · Answer

(A) Let $I = \int \frac{\cos^3 x}{\sin^2 x + \sin x} dx$.Substitute $\cos^2 x = 1 - \sin^2 x$:$I = \int \frac{(1 - \sin^2 x) \cos x}{\sin^2 x + \sin x} dx$.Let $u = \sin x$, then $du = \cos x dx$.$I = \int \frac{1 - u^2}{u^2 + u} du = \int \frac{(1 - u)(1 + u)}{u(u + 1)} du$.$I = \int \frac{1 - u}{u} du = \int (\frac{1}{u} - 1) du$.$I = \log|u| - u + c$.Substituting back $u = \sin x$, we get $I = \log|\sin x| - \sin x + c$.

$\int \frac{\cos^3 x}{\sin^2 x + \sin x} dx =$

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