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I am trying to solve this integral

$$\int_{0}^{\pi /2}\sin^n x\cdot dx.$$

I think we should solve it for:

a) odd numbers $2n+1$

$$\int_0^{\pi /2}\sin^{2n+1}x\cdot dx = \int_0^{\pi /2}\sin x\cdot \sin^{2n}x\cdot dx=\int_0^{\pi /2}\sin x\cdot  (1-cos^2x) ^n\cdot dx$$

let $t=\cos(x)$ and $dt=-\sin(x) \, dx$ then:

$$\int_0^{\pi /2}\sin x\cdot  (1-\cos^2x) ^n\cdot dx=-\int_1^0 (1-t^2)^n \, dt$$

Unfortunately I can not solve this integral. Would you please help me to finish it?

b) even numbers $2n$:

$$\int_0^{\pi /2}\sin^{2n}x\cdot dx = \int_0^{\pi /2}\left( \frac {1-\cos 2x} 2\right)^n \, dx$$

Unfortunately I can not solve this integral. Would you please help me to finish it?

I tried to search for something useful on the Internet and I found these two formulas: $$\int_0^{\pi /2}\sin^{2n+1}x\cdot dx = \int_0^{\pi /2}\cos^{2n+1}x\cdot dx = \frac{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}{1\cdot 3\cdot 5\cdot \ldots \cdot (2n+1)}\frac{\pi}{2}$$ $$\int_0^{\pi /2}\sin^{2n}x\cdot dx = \int_0^{\pi /2}\cos^{2n}x\cdot dx = \frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6 \cdot \ldots \cdot 2n}\frac{\pi}{2}$$

If you could write proofs of these two formulas that would solve my problem.

Thank you

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  • $\begingroup$ :for solving $$\int_0^{\pi /2}sinx\cdot (1-cos^2x) ^n\cdot dx=-\int_{1}^{0}(1-t^2)^ndt$$ use binomial formula $\endgroup$ – M.H Jun 24 '13 at 11:49
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – lab bhattacharjee Jun 24 '13 at 12:02
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You can do this by integration by parts, or equivalently using the product formula for differentiation. Let $I_n$ be the integral involving $\sin^n x$

So note that $$\frac {d}{dx}(\cos x\sin^{n-1}x)=-\sin^n x+(n-1)\cos^2x\sin^{n-2}x$$$$= -\sin^n x+(n-1)(1-\sin^2x)\sin^{n-2}x=-n\sin^nx+(n-1)sin^{n-2}x$$

Now integrate both sides, noting that: $\cos x\sin^{n-1}x=0$ at the limits of integration to obtain $$0=-nI_n+(n-1)I_{n-2}$$ which becomes $$I_n=\frac {(n-1)}nI_{n-2}$$

Using this successively gives the products you see in your answers, and takes you down to evaluating $I_1$ or $I_0$ (depending whether $n$ is odd or even). You should be able to finish it from there.

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  • $\begingroup$ May I ask why $(1-sin^2x)$ dissapeared on the left side? See $$-\sin^n x+(n-1)(1-\sin^2x)\sin^{n-2}x=-n\sin^nx+(n-1)sin^{n-2}x$$ $\endgroup$ – Anakin Jun 24 '13 at 16:03
  • $\begingroup$ $\sin^2x$ multiplied by $\sin^{n-2}x$ gives a term in $\sin^nx$ - which in full is $-(n-1)\sin^nx$. Then I gather the terms in $\sin^nx$ together to get the final expression. $\endgroup$ – Mark Bennet Jun 24 '13 at 16:11
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You can use recursive relation to get the integral. Let $$ I_n=\int_0^{\pi/2}\sin^nxdx. $$ Then using integration by parts, it is easy to get $$ I_n=\frac{n-1}{n}I_{n-2}. $$ So \begin{eqnarray*} I_{2n+1}=\frac{2n}{2n+1}I_{2n-1}=\cdots. \end{eqnarray*}

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