Evaluate the following integrals:

In this equation, we can manipulate numerator

$\sin ^{3} x=\sin ^{2} x \cdot \sin x$

$\therefore$ Now the equation becomes,

$\Rightarrow \int \frac{\sin ^{2} x \cdot \sin x}{\sqrt{\cos x}} d x$

$\sin ^{2} x=1-\cos ^{2} x$

$\Rightarrow \int \frac{1-\cos ^{2} x \cdot \sin x}{\sqrt{\cos x}} d x$

Now,

Let us assume $\cos x=t$

$\mathrm{d}(\cos x)=\mathrm{dt}$

$-\sin x d x=d t$

Substitute values of $\mathrm{t}$ and $\mathrm{dt}$ in above equation

$\Rightarrow-\int \frac{1-t^{2}}{\sqrt{t}} d t$

$\Rightarrow-\int \frac{1}{\sqrt{t}} d t-\int \frac{t^{2}}{\sqrt{t}} d t$

$\Rightarrow-\int t^{-1 \backslash 2} d t+\int t^{3} \backslash^{2} d t$

$\Rightarrow-2 t^{1 \backslash 2}+\frac{2}{5} t^{\frac{5}{2}}+c$

But $t=\cos x$

$\Rightarrow-2 \cos x^{1 \mid 2}+\frac{2}{5} \cos x^{\frac{5}{2}}+c$