# Evaluate the following integrals:

Question:

Evaluate $\int \sin ^{-1} \sqrt{x} d x$

Solution:

$\int \sin ^{-1} \sqrt{x} d x$

$\int u \cdot d v=u v-\int v d u$

Choose u in these order LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG,E-EXPONENTIAL)

$\mathrm{u}=\sin ^{-1} \sqrt{x} \mathrm{v}=1$

$\therefore \int \sin ^{-1} \sqrt{x}=\mathrm{x} \cdot \sin ^{-1} \sqrt{x}-\frac{1}{2} \int \frac{\sqrt{\mathrm{x}}}{\sqrt{1-x}} \mathrm{dx}$

Put $\sqrt{x}=t$

$\mathrm{d} \mathrm{x}=2 \mathrm{tdt}$

$=\mathrm{x} \cdot \sin ^{-1} \sqrt{x}-\int \frac{\mathrm{t}^{2}}{\sqrt{1-t^{2}}} \mathrm{dt}$

Now put $\mathrm{t}=$ sinu;

$\mathrm{dt}=\cos u \mathrm{du}$

$\sqrt{1-t^{2}}=\sqrt{1-\sin ^{2}} u$

$=\cos u$

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

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

$=\mathrm{x} \cdot \sin ^{-1} \sqrt{x}-\int \sin ^{2} u d u \ldots\left(\right.$ Here we can substitute $\left.\sin ^{2} \mathrm{x}=(1-\cos 2 \mathrm{u}) / 2\right)$

$=\mathrm{x} \cdot \sin ^{-1} \sqrt{x}-\int \frac{1-\cos 2 \mathrm{u}}{2} \mathrm{du}$

$=\mathrm{x} \cdot \sin ^{-1} \sqrt{x}-\left[\int \frac{1-\cos 2 \mathrm{u}}{2} \mathrm{du}\right]$

$=\mathrm{x} \cdot \sin ^{-1} \sqrt{x}-\left[\frac{u}{2}-\frac{1}{4} \sin 2 u\right]+c$

Put $u=\sin ^{-1} \sqrt{x}$

$\mathrm{I}=\mathrm{X} \cdot \sin ^{-1} \sqrt{x}-\left[\frac{\sin ^{-1} \sqrt{x}}{2}-\frac{\sqrt{x} \sqrt{(1-x)}}{2}\right]+C$