Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \sin ^{3} \sqrt{x} d x$

Solution:

Let

$\sqrt{X}=t$

$\mathrm{x}=\mathrm{t}^{2}$

$\mathrm{dx}=2 \mathrm{tdt}$

$I=2 \int t \sin ^{3} t d t$

$=2 \int \mathrm{t}\left(\frac{3 \sin t-\sin 3 t}{4}\right) \mathrm{dt}$

$=\frac{1}{2} \int \mathrm{t}(3 \sin t-\sin 3 \mathrm{t}) \mathrm{dt}$

Using integration by parts,

$=\frac{1}{2}\left[t\left(-3 \cos t+\frac{1}{3} \cos 3 t\right)-\int\left(-3 \cos t+\frac{\cos 3 t}{3}\right) d t\right]$

$=\frac{1}{2}\left[\frac{-9 t \cos t+t \cos 3 t}{3}-\left\{-3 \sin t+\frac{\sin 3 t}{9}\right\}\right]+c$

$=\frac{1}{2}\left[\frac{-9 \cos t+t \cos 3 t}{3}+\frac{27 \sin t-3 \sin 3 t}{9}\right]+c$

$=\frac{1}{18}[-27 \cos t+3 t \cos 3 t+27 \sin t-3 \sin 3 t]+c$

$I=\frac{1}{18}[3 \sqrt{x} \cos 3 \sqrt{x}+27 \sin \sqrt{x}-27 \sqrt{x} \cos \sqrt{x}-3 \sin 3 \sqrt{x}]+c$

Leave a comment

None
Free Study Material