Evaluate:

Question:

Evaluate: $\int \sin ^{3}(2 x+1) d x$

Solution:

We know $\sin 3 x=-4 \sin ^{3} x+3 \sin x$

$\Rightarrow 4 \sin ^{3} x=3 \sin x-\sin 3 x$

$\Rightarrow \sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4}$

$\Rightarrow \int \sin ^{3}(2 \mathrm{x}+1) \mathrm{d} \mathrm{x}=\int \frac{3 \sin (2 \mathrm{x}+1)-\sin 3(2 \mathrm{x}+1)}{4} \mathrm{dx}$

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

$\Rightarrow \frac{3}{8} \int \sin (2 x+1) d x-\frac{1}{4} \int \sin (6 x+3) d x$

$\Rightarrow \frac{-3}{8} \cos (2 x+1)+\frac{1}{24} \cos (6 x+3)+c$

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