The value of $\int \frac{\cos ^3 x}{\sin ^2 x+\sin x} \mathrm{~d} x$ is
The value of $\int \frac{\cos ^3 x}{\sin ^2 x+\sin x} \mathrm{~d} x$ is
$\quad \log (\sin x)-\sin x+\mathrm{c}$, where c is a constant of integration.
$\log (\sin x)-\cos x+\mathrm{c}$, where c is a constant of integration.
$\log (\sin x)+\sin x+\mathrm{c}$, where c is a constant of integration.
$\log (\cos x)-\cos x+\mathrm{c}$, where c is a constant of integration.
Solution
$\begin{aligned}
& I=\int \frac{\left(1-\sin ^2 x\right) \cos x}{\sin x(1+\sin x)} d x \\
& =\int \frac{1-\sin x}{\sin x} \cos x d x
\end{aligned}$
Put $t=\sin x$
$\Rightarrow \cos x d x=d t$
So, $I=\int\left(\frac{1}{t}-1\right) d t$
$\begin{aligned}
& =\log t-t+C \\
& \Rightarrow I=\log \sin x-\sin x+C
\end{aligned}$