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

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Question

(D) Let $I = \int \frac{\cos^3 x}{\sin^2 x + \sin x} \, dx$. Using the identity $\cos^2 x = 1 - \sin^2 x$,we can rewrite the integral as: $I = \int \f

Vedclass · Accepted Answer

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

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(D) Let $I = \int \frac{\cos^3 x}{\sin^2 x + \sin x} \, dx$. Using the identity $\cos^2 x = 1 - \sin^2 x$,we can rewrite the integral as: $I = \int \frac{(1 - \sin^2 x) \cos x}{\sin^2 x + \sin x} \, dx$. Substitute $\sin x = t$,which implies $\cos x \, dx = dt$. Substituting these into the integral: $I = \int \frac{1 - t^2}{t^2 + t} \, dt = \int \frac{(1 - t)(1 + t)}{t(t + 1)} \, dt$. Canceling the common term $(1 + t)$: $I = \int \frac{1 - t}{t} \, dt = \int \left( \frac{1}{t} - 1 \right) \, dt$. Integrating with respect to $t$: $I = \log |t| - t + C$. Substituting back $t = \sin x$: $I = \log |\sin x| - \sin x + C$.

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

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