Question:
Evaluate: $\int \sin ^{2}(2 x+5) d x$
Solution:
$\sin ^{2} x=\frac{1-\cos 2 x}{2}$
$\therefore$ The given equation becomes,
$\Rightarrow \int \frac{1-\cos 2(2 x+5)}{2} \mathrm{dx}$
We know $\int \cos a x d x=\frac{1}{a} \sin a x+c$
$\Rightarrow \frac{1}{2} \int \mathrm{dx}-\frac{1}{2} \int \cos (4 \mathrm{x}+10) \mathrm{dx}$
$\Rightarrow \frac{\mathrm{x}}{2}-\frac{1}{8} \sin (4 \mathrm{x}+10)+\mathrm{c}$
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