Question

Evaluate $\int \left[\sin \alpha \sin \right(x-\alpha )+{\sin }^2\left(\frac{x}{2}-\alpha \right)]dx$.

Answer 1

$I$ $=\int [\sin \alpha \sin (x-\alpha )+{\sin }^2(\frac{x}{2}-\alpha )]dx$

$I$ $=\sin \alpha \int \sin (x-\alpha )dx+\int {\sin }^2(\frac{x}{2}-\alpha )dx$

Putting $x-\alpha =p$ and $\frac{x}{2}-\alpha =q$

Thus, $dx=dp$ and $dx=2dq$

$I$ $=\sin \alpha \int \sin pdp+\int \left({\sin }^{2}q\right)2dq$

$I$ $=\sin \alpha (-\cos p)+\int \frac{1-\cos 2q}{2}\left(2dq\right)$

$I$ $=\sin \alpha (-\cos (x-\alpha ))+\int (1-\cos 2q)dq$

$I$ $=-\sin \alpha \cos (x-\alpha )+q-\frac{\sin 2q}{2}+C$

$I$ $=-\sin \alpha \cos (x-\alpha )+(\frac{x}{2}-\alpha )-\sin (\frac{x}{2}-\alpha )\cos (\frac{x}{2}-\alpha )+C$