Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{x^{3} \sin ^{-1} x^{2}}{\sqrt{1-x^{4}}} d x$

Solution:

Let $I=\int \frac{x^{3} \sin ^{-1} x^{2}}{\sqrt{1-x^{4}}} d x$

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

$\frac{1}{\sqrt{1-x^{4}}} 2 x d x=d t$

$I=\int \frac{x^{2} \sin ^{-1} x^{2}}{\sqrt{1-x^{4}}} x d x$

$=\int(\sin t) t \frac{d t}{2}$

Using integration by parts,

$=\frac{1}{2}\left[\mathrm{t} \int \sin t \mathrm{~d} t-\int \frac{\mathrm{d}}{\mathrm{dt}} \mathrm{t} \int \sin t \mathrm{dt}\right]$

$=\frac{1}{2}\left[-\mathrm{t} \cos t-\int-\mathrm{costdt}\right]$

$=\frac{1}{2}[-t \cos t+\sin t]+c$

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

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now