Question:
$\int \frac{\sin -}{\sin -} d x$ is equal to :
(where $c$ is a constant of integration.)
Correct Option: , 3
Solution:
$\int \frac{\sin \left(\frac{5 x}{2}\right)}{\sin \left(\frac{x}{2}\right)} d x=\int \frac{2 \cos \frac{x}{2} \cdot \sin \frac{5 x}{2}}{2 \cos \frac{x}{2} \cdot \sin \frac{x}{2}} d x$
$=\int \frac{\sin 3 x+\sin 2 x}{\sin x} d x$
$=\int\left(3-4 \sin ^{2} x+2 \cos x\right) d x$
$\left[\because \sin 2 x=2 \sin x \cos x\right.$ and $\left.\sin 3 x=3 \sin x-4 \sin ^{3} x\right]$
$=\int(3-2(1-\cos 2 x)+2 \cos x) d x$
$=\int(1+2 \cos x+2 \cos 2 x) d x$
$=x+2 \sin x+\sin 2 x+c$