Thread: Trig Substitution
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Old October 2nd, 2007, 10:35 AM
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\displaystyle I_n=\int\sin^nxdx=\int\sin^{n-1}x\sin xdx=\int\sin^{n-1}x(-\cos x)'dx=
\displaystyle=-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}\cos^2dx=
\displaystyle=-\sin^{n-1}x\cos x+(n-1)\int(\sin^{n-2}x-\sin^nx)dx=
\displaystyle=-\sin^{n-1}x\cos x+(n-1)I_{n-2}-(n-1)I_n

Then I_n=-\frac{1}{n}\sin^{n-1}x\cos x+\frac{n-1}{n}I_{n-2}
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