Evaluate the following integrals:
$\int \frac{\cos ^{3} x}{\sqrt{\sin x}} d x$
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
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