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I am in war with my textbook. It says that $\int \sin^n{x}\cos^m{x}\ dx=\frac{\sin^{n+1}x\cos^{m-1}x}{n+m}+\frac{m-1}{n+m}\int \sin{x}\cos^{m-2}\ dx$.

Then I wrote the proof. Here is what I got. \begin{equation*} \begin{split} \int \sin^n{x}\cos^m{x}\ dx&=\int \sin^n{x}\cos^{m-1}{x}\cos{x}\ dx\\ &=\int \sin^{n}{x}\cos^{m-1}\ d(\sin{x})\\ &=\cos^{m-1}x\frac{\sin^{n+1}{x}}{n+1}-\int \frac{\sin^{n+1}{x}}{n+1}(m-1)\cos^{m-2}{x}\ dx\\ &=\frac{\sin^{n+1}{x}\cos^{m-1}{x}}{n+1}-\frac{m-1}{n+1}\int\sin^{n+1}{x}\cos^{m-2}{x}\ dx \end{split} \end{equation*}

Maybe you know what I missed?

naomi72
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1 Answers1

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I think your miss is here:$$\int \sin ^nx\cos ^{m-1}xd(\sin x)=\cos ^{m-1}x\frac{\sin ^{n+1}x}{n+1}-\int \frac{\sin ^{n+1}x}{n+1}(m-1)\cos ^{m-2}xd({\bf cos x})$$ and similarly as you have done, can use $$\int \sin ^nx\cos ^mxdx=\frac{1}{m+n}\int \sin ^{n-(m+n)+1}x\cos ^{m-1}xd(\sin ^{m+n}x)=\frac{\sin ^{n+1}x\cos ^{m-1}x}{m+n}-\frac{1}{m+n}\int \sin ^{m+n}x \left((1-m)\sin ^{-m}x\cos ^mx-(m-1)\sin ^{2-m}x\cos ^{m-2}x\right)dx=\frac{\sin ^{n+1}x\cos ^{m-1}x}{m+n}+\frac{m-1}{m+n}\int \sin ^nx\cos ^{m-2}(\sin^2x+\cos^2x)xdx$$ so I think there should be a $\sin ^n$ in the second term of the r.h.s

Alexis
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