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Does it hold? $$\int_{0}^{\Large\frac\pi2}\cos^{2k}x\ dx=\frac{(2k-1)!!}{2k!!}\cdot\frac{\pi}{2}$$ where $k \in \mathbb Z$

I have seen some examples like $$\int_{0}^{\Large\frac\pi2}\cos^{6}x\ dx=\frac{5 \cdot 3}{6 \cdot 4 \cdot 2}\cdot\frac{\pi}{2}$$ and I conclude something like this.

I try to use Taylor series but it doesn't work well. Do you have good idea? Thank you a lot!

Seyhmus Güngören
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Zhen Zhang
  • 805

2 Answers2

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Let $$ I_n=\int\cos^n x\ dx. $$ Using integration by reduction formula, we will obtain \begin{align} \int\cos^nx\ dx&=\frac1n\cos^{n-1}x\sin x+\frac{n-1}{n}\int\cos^{n-2}x\ dx\\ I_n&=\frac1n\cos^{n-1}x\sin x+\frac{n-1}{n}I_{n-2}. \end{align} It turns out that $$ \int_0^{\Large\frac\pi2}\cos^nx\ dx=\frac{n-1}{n}\int_0^{\Large\frac\pi2}\cos^{n-2}x\ dx.\tag1 $$ Applying $(1)$ for several times, we will obtain the pattern that leads to the result $$ \int_0^{\Large\frac\pi2}\cos^nx\ dx=\large\color{green}{\frac{(n-1)!!}{n!!}\cdot\frac\pi2},\tag2 $$ where $I_0=\dfrac\pi2$.

Setting $n=2k$ to $(2)$ yields $$ \int_0^{\Large\frac\pi2}\cos^{2k}x\ dx=\large\color{blue}{\frac{(2k-1)!!}{(2k)!!}\cdot\frac\pi2}. $$

Tunk-Fey
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Another approach:

Consider Beta function $$ \text{B}(u,v)=2\int_0^{\Large\frac\pi2}(\sin x)^{2u-1}(\cos x)^{2v-1}\ dx=\frac{\Gamma(u)\cdot\Gamma(v)}{\Gamma(u+v)}, $$ then $$ \int_0^{\Large\frac\pi2}\cos^{2k}x\ dx=\frac{\Gamma\left(\dfrac12\right)\cdot\Gamma\left(k+\dfrac12\right)}{2\cdot\Gamma\left(k+1\right)}=\large\color{blue}{\frac{(2k-1)!!}{(2k)!!}\cdot\frac\pi2}, $$ where

  • $\Gamma(k)=k!$
  • $\Gamma\left(k+\dfrac12\right)=\dfrac{(2k-1)!!}{2^k}\sqrt{\pi}$
  • $(2k)!!=2^k\cdot k!$
  • $\Gamma\left(\dfrac12\right)=\sqrt\pi$

for $k=0,1,2,3,\cdots$.

The proof of above identities can be seen here and here.

Tunk-Fey
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