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If $I_n=\int _0^{\pi }\:sin^{2n}\theta \:d\theta $, show that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, and hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $

Hence calculate $\int _0^{\pi }\:\:sin^4tcos^6t\:dt$

I knew how to prove that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$ ,, but I am not very good at English, what does it mean Hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $ do we need to prove this part as well or is it just a hint to use? and for the other calculation to find $\int _0^{\pi }\:\:sin^4tcos^6t\:dt$ is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question.

ASD123
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ASD123
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4 Answers4

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Hint

Integrate by parts

$$I_n=\int_0^\pi \sin\theta\sin^{2n-1}\theta d\theta$$ and use the identity

$$\cos^2\theta+\sin^2\theta=1$$

  • I didn't get why we use the Identity as we still don't have sin^2 ? can you please explain it more simply and show more steps – ASD123 Nov 17 '14 at 20:15
  • $$I_n=-\cos\theta \sin^{2n-1}\theta\bigg|0^\pi+(2n-1)\int_0^\pi\sin^{2n-2}\theta\cos^2\theta d\theta$$ and with the mentioned identity we get $$I_n=(2n-1)(I{n-2}-I_n)$$ –  Nov 17 '14 at 20:24
  • I am sorry but I didn't understand any part of the answer,, where did u get the cos form at the beginning, did u use any other integration method like integration by part or something else? if it's possible can u try to solve it entirely step by step with mentioning any method used on the steps I would be grateful for you sir. Thank you very much for helping – ASD123 Nov 17 '14 at 20:29
  • Let $u=\sin^{2n-1}\theta$ then $u'=(2n-1)\sin^{2n-2}\theta\cos\theta$ and let $v'=\sin\theta$ so $v=-\cos\theta$ hence by integration by parts we have $$\int_0^\pi v'u d\theta= uv\bigg|_0^\pi-\int_0^\pi vu' d\theta$$ –  Nov 17 '14 at 20:35
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My answer it attached in two pages

integrationByPartsFormula

Idris Addou
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Here is the last part of the problem you post. You can verify the value of the definite integral in the wolfram, at

http://www.wolframalpha.com/input/?i=integral_0^%28pi%29+sin^4%28x%29cos^6%28x%29dx

Last part

Idris Addou
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what does it mean Hence $I_n = \frac{(2n)!\pi}{(2^n n!)2}$ do we need to prove this part as well or is it just a hint to use?

"Hence" means "from here"; it is an archaic word. It survives in mathematical English, where it means "because of what was just said", or "using what was just proved". They're saying that you should use $I_n = \frac{2n-1}{2n} I_{n-1}$ to prove $I_n = \frac{(2n)!\pi}{(2^n n!)2}$.