Find the following integral:int sin ^{3} x co | vedclass

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(A) We have $\int \sin ^{3} x \cos ^{2} x \, dx = \int \sin ^{2} x \cos ^{2} x (\sin x) \, dx$$= \int (1 - \cos ^{2} x) \cos ^{2} x (\sin x) \, dx$Let

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$-\frac{1}{3} \cos ^{3} x + \frac{1}{5} \cos ^{5} x + C$

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(A) We have $\int \sin ^{3} x \cos ^{2} x \, dx = \int \sin ^{2} x \cos ^{2} x (\sin x) \, dx$$= \int (1 - \cos ^{2} x) \cos ^{2} x (\sin x) \, dx$Let $t = \cos x$,then $dt = -\sin x \, dx$,which implies $\sin x \, dx = -dt$Substituting these into the integral,we get:$= -\int (1 - t^{2}) t^{2} \, dt$$= -\int (t^{2} - t^{4}) \, dt$$= -(\frac{t^{3}}{3} - \frac{t^{5}}{5}) + C$$= -\frac{1}{3} t^{3} + \frac{1}{5} t^{5} + C$Substituting $t = \cos x$ back,we get:$= -\frac{1}{3} \cos ^{3} x + \frac{1}{5} \cos ^{5} x + C$

Find the following integral:
$\int \sin ^{3} x \cos ^{2} x \, dx$

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Find the following integral:$\int \sin ^{3} x \cos ^{2} x \, dx$