Evaluate the following integrals:

In this equation, we can manipulate numerator

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

$\therefore$ Now the equation becomes,

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

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

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

Now,

Let us assume $\sin x=t$

$d(\sin x)=d t$

$\cos x d x=d t$

Substitute values of $t$ and $d t$ in the 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 $t=\sin x$

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

RD Sharma Solutions for Class 12 Math Chapter 14 - Indefinite Integrals