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Question Number 114467 by Eric002 last updated on 19/Sep/20

$$\int x{sin}^n\left(x\right)dx$$

Answered by Olaf last updated on 19/Sep/20

$${\mathrm{I}}_n\left(x\right)=\int x{\sin }^{n}xdx$$$${\mathrm{I}}_n\left(x\right)-{\mathrm{I}}_{n+2}\left(x\right)=\int x{\sin }^{n}x(1-{\sin }^{2}x)dx$$$${\mathrm{I}}_n\left(x\right)-{\mathrm{I}}_{n+2}\left(x\right)=\int \left(x\cos x\right)\left(\cos x{\sin }^{n}x\right)dx$$$$u=x\cos x,u^{\prime }=\cos x-x\sin x$$$$v^{\prime }=\cos x{\sin }^{n}x,v=\frac{{\sin }^{n+1}x}{n+1}$$$${\mathrm{I}}_n\left(x\right)-{\mathrm{I}}_{n+2}\left(x\right)=\frac{1}{n+1}x\cos x{\sin }^{n+1}x$$$$-\int (\cos x-x\sin x)\frac{{\sin }^{n+1}x}{n+1}dx$$$${\mathrm{I}}_n\left(x\right)-{\mathrm{I}}_{n+2}\left(x\right)=\frac{1}{n+1}x\cos x{\sin }^{n+1}x$$$$-\frac{1}{n+1}\int \cos x{\sin }^{n+1}xdx+\frac{1}{n+1}\int x{\sin }^{n+2}xdx$$$${\mathrm{I}}_n\left(x\right)-{\mathrm{I}}_{n+2}\left(x\right)=\frac{1}{n+1}x\cos x{\sin }^{n+1}x$$$$-\frac{1}{(n+1)(n+2)}{\sin }^{n+2}x+\frac{1}{n+1}{\mathrm{I}}_{n+2}\left(x\right)$$$$$$$$\frac{n+2}{n+1}{\mathrm{I}}_{n+2}\left(x\right)-{\mathrm{I}}_n\left(x\right)=\frac{{\sin }^{n+1}x}{n+1}(\frac{\sin x}{n+2}-x\cos x)$$$${\mathrm{I}}_{n+2}\left(x\right)=\frac{n+1}{n+2}{\mathrm{I}}_n\left(x\right)+\frac{{\sin }^{n+1}x}{n+2}(\frac{\sin x}{n+2}-x\cos x)$$$$\mathrm{and}{\mathrm{I}}_0\left(x\right)=\int xdx=\frac{x^2}{2}$$$${\mathrm{I}}_1\left(x\right)=\int x\sin xdx=\sin x-x\cos x$$$$\mathrm{work}\mathrm{in}\mathrm{progress}...$$

Commented by Eric002 last updated on 19/Sep/20

$$welldone$$

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