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Q.
The value of $\int \frac{\cos ^3 x}{\sin ^2 x+\sin x} d x$ is equal to:
Integrals
Solution:
$I=\int \frac{\cos ^3 x}{\sin ^2 x+\sin x} d x=\int \frac{\cos x \cdot\left(1-\sin ^2 x\right)}{\sin x(1+\sin x)} d x$
Put $\sin x=t$, then $\cos x d x=d t$
$\Rightarrow I=\int \frac{(1-t)(1+t) d t}{t(1+t)}$
$=\ell n|t|-t+C=\ell n|\sin x|-\sin x+C$