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I've found online that $\int_0^{\frac{\pi}{2}}\cos^{2n}{(x)}\,dx = \frac{\pi}{2^{2n-1}} \binom{2n}{n}$. I've tried it on my own and I've got $\int_0^{\frac{\pi}{2}}\cos^{2n}{x}\,dx = \frac{\pi}{2^{n+1}} \frac{(2n-1)!!}{n!}$, where the double factorial is the product of all odd numbers from $1$ up to $2n-1$.

To compute the integral, I've integrated by parts splitting $\cos^{2n}{(x)}$ in $\cos^{2n-1}{(x)}$ and $\cos{(x)}$, then iterating, obtaining the following:

$I_n = \int_0^{\frac{\pi}{2}}\cos^{2n}{(x)}\,dx = (2n-1)\int_0^{\frac{\pi}{2}}(\cos^{2n-2}{(x)} - \cos^{2n}{(x)})\,dx = (2n-1)(I_{n-1} - I_n)$

$\Rightarrow I_n = \frac{2n-1}{2n}I_{n-1} \Rightarrow I_n = \frac{(2n-1)!!}{2^n n!} I_0 = \frac{(2n-1)!!}{2^n n!} \int_0^{\frac{\pi}{2}}\,dx = \frac{\pi}{2^{n+1}} \frac{(2n-1)!!}{n!}$

Unfortunately for me, this formula is wrong, but I can't quite figure out where I'm going wrong.

Could somebody please help me find the mistake?

RobPratt
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Massi Piero
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2 Answers2

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Note that \begin{align}\int_0^{\frac{\pi}{2}}\cos^{2n}{x}\,dx =&\ \frac{\pi}{2^{2n+1}} \binom{2n}{n} =\frac{\pi}{2^{2n+1}} \frac{(2n)!}{(n!)^2}\\ =&\ \frac{\pi}{2^{2n+1}} \frac{(2n)!!(2n-1)!!}{(n!)^2} =\frac{\pi}{2^{2n+1}} \frac{2^n n!(2n-1)!!}{(n!)^2}\\ =&\ \frac{\pi}{2^{n+1}} \frac{(2n-1)!!}{n!} \end{align} So, your result is correct.

Quanto
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First of all, you cited the online source incorrectly by a factor of $4$. In fact, the source states your integral $$I = \int_0^{\pi/2} \cos^{2n}x dx = \frac{\pi}{2^{2n+1}}\binom{2n}{n}=:(\ast)$$

But this is the same as your calculation as $(2n-1)!!(2n)!!=(2n)!$ so $(2n-1)!!=\frac{(2n)!}{2^nn!}$, and thus your calculation

$$=\frac{\pi}{2^{n+1}}\frac{(2n-1)!!}{n!}=\frac{\pi}{2^{n+1}2^n}\frac{(2n)!}{n!n!}=(\ast)=I.$$

Benjamin Wang
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