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

Evaluate:

Question:

Evaluate: $\int \sin ^{2} \frac{x}{2} d x$

Solution:

$\sin ^{2} x=\frac{1-\cos 2 x}{2}$

$\therefore$ The given equation becomes,

$\Rightarrow \int \frac{1-\cos 2 \frac{x}{2}}{2} \mathrm{dx}=\int \frac{1-\cos x}{2} \mathrm{dx}$

We know $\int \cos a x d x=\frac{1}{a} \sin a x+c$

$\Rightarrow \frac{1}{2} \int d x-\frac{1}{2} \int \cos (x) d x$

$\Rightarrow \frac{x}{2}-\frac{1}{2} \sin (x)+c$

Leave a comment

None
Free Study Material