Write integral of sin as a multiple

Question

Our professor gave us this question.

Write $\int^{2\pi}_{0}sin^{100}(x)dx$ as a multiple of $\int^{2\pi}_{0}sin^{98}(x)dx$ using simple techniques (like substitution or integration by parts). Can somebody give me a start on how to go about it?

There are formulas for reduction of powers that are available in most calculus textbooks. That use the same method as Khosrotash employed but in a more general setting. — Tucker, Aug 02 '15 at 02:57

score 1 · Answer 1 · answered Aug 02 '15 at 00:53

1

Remember: $ \sin^{100}(x)= \sin^{98}(x) \sin^{2}(x)= sin^{98}(x) (1-cos^{2}(x))$

answered Aug 02 '15 at 00:53

Race Bannon

1,829

This doesn't really answer the question. This lets you express it in terms of the integral of $\sin^{98}x$ and the integral of $\sin^{98}x \cos^2x$, which is not the goal. – Matt Samuel Aug 02 '15 at 03:56

score 1 · Accepted Answer · edited Aug 02 '15 at 01:27

Use integration by part $$I_{100}=\int^{2\pi}_{0}\sin^{100}x\,dx=\\\int^{2\pi}_{0}(\sin x)^{99}\cdot{\color{Green} {\sin x}} \,dx\\ =({\color{Green} {-\cos x}}) \sin^{99}x \Big|_{0}^{2\pi}-\int^{2\pi}_{0}(99(\sin x)^{98}\cdot\cos x )({\color{Green} {-\cos x}})\, dx=\\ 0 -(-99)\int^{2\pi}_{0}\sin^{98}x\cdot(\cos^2x)=\\99\int^{2\pi}_{0}\sin^{98}x(1-\sin^2x) dx=\\99\int^{2\pi}_{0}(\sin^{98}x-\sin^{100}x)dx\\=I_{100}=99\left(\int^{2\pi}_{0}(\sin^{98}x)dx-\int^{2\pi}_{0}\sin^{100}x\,dx\right) \to$$ $$ {\color{Red}{I_{100}=99(I_{98}-I_{100})} }\\100I_{100}=99I_{98}\\$$

Write integral of sin as a multiple

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