Question:
$\sin ^{2}(2 x+5)$
Solution:
$\sin ^{2}(2 x+5)=\frac{1-\cos 2(2 x+5)}{2}=\frac{1-\cos (4 x+10)}{2}$
$\Rightarrow \int \sin ^{2}(2 x+5) d x=\int \frac{1-\cos (4 x+10)}{2} d x$
$=\frac{1}{2} \int 1 d x-\frac{1}{2} \int \cos (4 x+10) d x$
$=\frac{1}{2} x-\frac{1}{2}\left(\frac{\sin (4 x+10)}{4}\right)+\mathrm{C}$
$=\frac{1}{2} x-\frac{1}{8} \sin (4 x+10)+\mathrm{C}$
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