The value of int frac{cos ^3 x}{sin ^2 x+sin | vedclass

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Question

(A) Let $I = \int \frac{\cos ^3 x}{\sin ^2 x+\sin x} \,d x$. Using the identity $\cos^2 x = 1 - \sin^2 x$, we have: $I = \int \frac{(1-\sin^2 x) \cos

Vedclass · Accepted Answer

$\log (\sin x)-\sin x+C$, where $C$ is a constant of integration.

Vedclass · Answer

(A) Let $I = \int \frac{\cos ^3 x}{\sin ^2 x+\sin x} \,d x$. Using the identity $\cos^2 x = 1 - \sin^2 x$, we have: $I = \int \frac{(1-\sin^2 x) \cos x}{\sin x(1+\sin x)} \,d x$. Since $1-\sin^2 x = (1-\sin x)(1+\sin x)$, the expression simplifies to: $I = \int \frac{(1-\sin x)(1+\sin x) \cos x}{\sin x(1+\sin x)} \,d x = \int \frac{1-\sin x}{\sin x} \cos x \,d x$. Substitute $t = \sin x$, then $dt = \cos x \,d x$. $I = \int \left(\frac{1}{t} - 1\right) dt$. Integrating with respect to $t$: $I = \log |t| - t + C$. Substituting back $t = \sin x$: $I = \log |\sin x| - \sin x + C$.

The value of $\int \frac{\cos ^3 x}{\sin ^2 x+\sin x} \,d x$ is

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