Evaluate the following integrals:


Evaluate the following integrals:

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


Let $I=\sin ^{-1} \sqrt{x} d x$

$\sqrt{\mathrm{X}}=\mathrm{t} ; \mathrm{x}=\mathrm{t}^{2}$

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

$=\sin ^{-1} \mathrm{t} 2 \mathrm{t} \mathrm{dt}$

Using integration by parts,

$=\sin ^{-1} t \int 2 t d t-\int \frac{d}{d t} \sin ^{-1} t \int 2 t d t$

We know that, $\frac{\mathrm{d}}{\mathrm{dt}} \sin ^{-1} \mathrm{t}=\frac{\mathrm{t}}{\sqrt{1-\mathrm{t}^{2}}}$

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

let us solve, $\int \frac{t^{2}}{\sqrt{1-t^{2}}} d t$

$=\int \frac{\mathrm{t}^{2}-1+1}{\sqrt{1-\mathrm{t}^{2}}} \mathrm{dt}=\int \frac{\mathrm{t}^{2}-1}{\sqrt{1-\mathrm{t}^{2}}} \mathrm{dt}+\int \frac{1}{\sqrt{1-\mathrm{t}^{2}}} \mathrm{~d} \mathrm{t}$

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

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

$\mathrm{t}=\sin \mathrm{u} ; \mathrm{dt}=\cos \mathrm{u} \mathrm{du}$

$\int-\sqrt{1-\mathrm{t}^{2}} \mathrm{dt}=\int-\cos ^{2} \mathrm{u} \mathrm{d} \mathrm{u}=-\int\left[\frac{1+\cos 2 \mathrm{u}}{2}\right] \mathrm{du}$

$=-\frac{u}{2}-\frac{\sin 2 u}{4}$

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

$=-\frac{\sin ^{-1} t}{2}-\frac{\sin \left(2 \sin ^{-1} t\right)}{4}$

There fore, $\int \sin ^{-1} \sqrt{x} d x=x \sin ^{-1} \sqrt{x}-\frac{\sin ^{-1} \sqrt{x}}{2}-\frac{\sin \left(2 \sin ^{-1} t\right)}{4}$

$=x \sin ^{-1} \sqrt{x}-\frac{\sin ^{-1} \sqrt{x}}{2}-\frac{\sqrt{x(1-x)}}{2}$

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