Evaluate the following integrals:

Solution:

Let, $x=\sin ^{2} t$

Differentiate both side with respect to t

$\frac{d x}{d t}=2 \sin t \cos t d t \Rightarrow \mathrm{dx}=2 \sin \mathrm{t} \cos \mathrm{tdt}$

$y=\int \frac{\sin ^{4} t}{\cos t} 2 \sin t \cos t d t$

$y=2 \int \sin ^{5} t d t$

$y=2 \int\left(1-\cos ^{2} t\right)^{2} \sin t d t$

Let, $\cos t=z$

Differentiate both side with respect to $z$

$-\sin t \frac{d t}{d z}=1 \Rightarrow \sin \mathrm{tdt}=-\mathrm{dz}$

$y=-2 \int\left(1-z^{2}\right)^{2} d z$

$y=-2 \int 1+z^{4}-2 z^{2} d z$

$y=-2\left(z+\frac{z^{5}}{5}-2 \frac{z^{3}}{3}\right)+c$

Again put $z=\cos t$ and $t=\sin ^{-1} \sqrt{x}$

$y=-2\left(\cos \left(\sin ^{-1} \sqrt{x}\right)+\frac{\cos ^{5}\left(\sin ^{-1} \sqrt{x}\right)}{5}-2 \frac{\cos ^{3}\left(\sin ^{-1} \sqrt{x}\right)}{3}\right)+c$

$y=-2\left(\sqrt{1-x}+\frac{(1-x)^{2} \sqrt{1-x}}{5}-\frac{2(1-x) \sqrt{1-x}}{3}\right)+c$

$y=\frac{-2}{15} \sqrt{1-x}\left(3 x^{2}+4 x+8\right)+c$