Find the integral of the function sin ^{2}(2 | vedclass

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Question

(A) We know that $\sin ^{2} 	heta = \frac{1 - \cos 2	heta}{2}$.Substituting $	heta = 2x + 5$,we get:$\sin ^{2}(2 x+5) = \frac{1 - \cos 2(2 x+5)}{2}

Vedclass · Accepted Answer

$\frac{1}{2} x - \frac{1}{8} \sin (4 x + 10) + C$

Vedclass · Answer

(A) We know that $\sin ^{2} 	heta = \frac{1 - \cos 2	heta}{2}$.Substituting $	heta = 2x + 5$,we get:$\sin ^{2}(2 x+5) = \frac{1 - \cos 2(2 x+5)}{2} = \frac{1 - \cos (4 x+10)}{2}$.Now,we integrate the expression:$\int \sin ^{2}(2 x+5) dx = \int \frac{1 - \cos (4 x+10)}{2} dx$$= \frac{1}{2} \int 1 dx - \frac{1}{2} \int \cos (4 x+10) dx$$= \frac{1}{2} x - \frac{1}{2} \left( \frac{\sin (4 x+10)}{4} ight) + C$$= \frac{1}{2} x - \frac{1}{8} \sin (4 x+10) + C$.

Find the integral of the function $\sin ^{2}(2 x+5)$.

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