Integration - 1996 HSC 3b (1 Viewer)

AAEldar · Aug 7, 2011

$\text{By writing }(cosx)^{2n}=(1-sin^2x)^n\text{, show that }\int_{0}^{\frac{\pi}{2}}(cosx)^{2n+1}dx=\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\binom{n}{k}$

I can see why it would equal that, but I just can't get there >.< Probably something obvious but I'm just not seeing it.

ohexploitable · Aug 7, 2011

$\\\int_0^{\frac{pi}{2}}(1-\sin^2x)^n\cosx\,dx$

let u=sinx

$\\=\int_0^1(1-u^2)^n\,dx\\=\int_0^1(1-\binom{n}{1}u^2+\binom{n}{2}u^4-\binom{n}{3}u^6+...+\binom{n}{n}u^{2n})\,dx\\=1-\frac{1}{3}\binom{n}{1}+\frac{1}{5}\binom{n}{2}-\frac{1}{7}\binom{n}{3}+...+\frac{1}{2n+1}\binom{n}{n}\\=\sum_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}{k}$

deterministic · Aug 7, 2011

$\int_0^{\frac{\pi}{2}} (cosx)^{2n}cosxdx &= \int_0^{\frac{\pi}{2}} (1-sin^2x)^n cosxdx\\&=\int_0^1 (1-u^2)^n du \\&=\int_0^1 \sum_{k=0}^n ^nC_k (-u^2)^k du \\&=\sum_{k=0}^n ^nC_k \left[\frac{(-1)^ku^{2k+1}}{2k+1}\right]_0^1\\&=\sum_{k=0}^n ^nC_k \frac{(-1)^k}{2k+1}$

edit: see above ^^

AAEldar · Aug 7, 2011

EDIT:

LOLOL don't worry hahaha.

Thanks for the help guys!

Drongoski · Aug 7, 2011

ohexploitable said:
$\\\int_0^{\frac{pi}{2}}(1-\sin^2x)^n\cosx\,dx$

let u=sinx

$\\=\int_0^1(1-u^2)^n\,dx\\=\int_0^1(1-\binom{n}{1}u^2+\binom{n}{2}u^4-\binom{n}{3}u^6+...+\binom{n}{n}u^{2n})\,dx\\=1-\frac{1}{3}\binom{n}{1}+\frac{1}{5}\binom{n}{2}-\frac{1}{7}\binom{n}{3}+...+\frac{1}{2n+1}\binom{n}{n}\\=\sum_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}{k}$

Just a minor typo: subsequent dx should read du.

Integration - 1996 HSC 3b (1 Viewer)

AAEldar

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ohexploitable

hey

deterministic

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AAEldar

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Drongoski

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