Calc1.62

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\int \sin ^{2}x\cos ^{3}x\,dx\,

Since the power of \cos x is odd, turn all but one into \sin x and do a substitution. Let

I=\int \sin ^{2}x\cos ^{3}x\,dx=\int \sin ^{2}x(1-\sin ^{2}x)\cos x\,dx\,

Let u=\sin x and du=\cos xdx. So

I=\int u^{2}(1-u^{2})\,du=\int u^{2}-u^{4}\,du={\frac  {1}{3}}u^{3}-{\frac  {1}{5}}u^{5}+C\,

And back substituting gives

\int \sin ^{2}x\cos ^{3}x\,dx={\frac  {1}{3}}\sin ^{3}x-{\frac  {1}{5}}\sin ^{5}x+C\,


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