Trigonometric Series Proof

Question

I am posed with the following question:

Prove that for even powers of $\sin$:

$$ \int_0^{\pi/2} \sin^{2n}(x) dx = \dfrac{1 \cdot 3 \cdot 5\cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} \times \dfrac{\pi}{2} $$

Here is my work so far:

1. Proof by induction

$P(1) \Rightarrow n = 2 \Rightarrow $

$$\int_0^{\pi/2} \sin^4(x) dx = \dfrac{1 \cdot 3}{2 \cdot 4} \times \dfrac{\pi}{2} $$

$$ \frac{3\pi}{16} = \frac{3 \pi}{16} $$

Base Case Succeeds. (The left hand side evaluated by MAPLE)

$P(n+1) \Rightarrow n = n + 1 \Rightarrow$

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

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

Let $u = \cos x, du = - \sin x dx \Rightarrow$

$$ -\int_0^{\pi/2} (1-u^2)^n du$$

Now I am stuck and unsure what to do with this proof ... Any help would be greatly appreciated

(I also tried using the reduction formulas to no avail)

EDIT

I have completed the proof.

I am posting this here for any other people who may have the same question...

We will prove by induction that $\forall n \in 2 \mathbb{N}_{>0}$ \begin{align*} \int_0^{\pi/2} \sin^{2n} x dx & = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{2 \times 4 \times 6 \times \cdots \times 2n} \times \frac{\pi}{2} \tag{1} \end{align*} With $k = 1$ as our base case, we have \begin{align*} \frac{1}{2}x - \frac{1}{4} \sin{2x} \bigg|_{0}^{\pi / 2} & = \frac{\pi}{4} \tag{67} \\ \frac{1}{2} \times \frac{\pi}{2} & = \frac{\pi}{4} \end{align*} Let $n \in 2\mathbb{N}_{>0}$ be given and suppose (1) is true for $k = n$. Then \begin{align*} \int_0^{\pi/2} \sin^{2n+2} x dx & = - \frac{1}{2n+2} \cos^{2n-1}x \sin x \bigg|_{0}^{\pi / 2} + \frac{2n+1}{2n+2} \int_0^{\pi/2} \sin^{2n} x dx \tag{67} \\ & = \frac{2n+1}{2n+2} \int_0^{\pi/2} \sin^{2n} x dx \\ & = \frac{1 \times 3 \times 5 \times \cdots \times (2n + 1)}{2 \times 4 \times 6 \times \cdots \times {(2n+2)}} \times \frac{2n+2}{2n+1} \times \frac{\pi}{2} \\ & = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{2 \times 4 \times 6 \times \cdots \times 2n} \times \frac{\pi}{2} \\ & = \int_0^{\pi/2} \sin^{2n} x dx \end{align*}

Thus, (1) holds for $k = n + 1$, and by the principle of induction, it follows that that (1) holds for all even numbers. $\square$

Answer 1

This proof makes use of the gamma and beta functions. We have the basic identity $$B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} $$

Note that $$B(x, y) = \int_0^1 t^x(1 - t)^y \,dt $$

by definition. Put $x = \sin^2\theta$. Thus we have $$\int_{0}^{\pi/2}\sin^{2m + 1}\theta\cos^{2n + 1}\theta \,\,d\theta = \frac{\Gamma(m)\Gamma(n)}{2\Gamma(m + n)}$$

Note that since we were originally dealing with two independent variables, we need to maintain those after making this substitution, hence the $m$ and $n$. Now put $2k = 2m + 1$ and $2n + 1 = 0$. This gives the following result:

$$\int_{0}^{\pi/2}\sin^{2k}\theta \,d\theta = \frac{\Gamma[\frac{1}{2}(2k + 1)]\Gamma(1/2)}{2\Gamma(k + 1)} $$

The denominator is simply $2\cdot k!$ and the numerator is $\pi\cdot k!\cdot \binom{k - 1/2}{k}$

Expanding the binomial coefficient gives the required result.

Answer 2

For $k=1$, it's straightforward to verify$$\int_0^{\pi/2}\sin^2x~dx=\int_0^{\pi/2}\frac{1-\cos 2x}2dx=\frac\pi4$$ Assume $k=n$ we have $$I_n=\int_0^{\pi/2}\sin^{2n}x~dx=\frac{(2n-1)!!}{(2n)!!}\frac\pi2$$ Then for $k=n+1$, $$\begin{align}I_{n+1}&=\int_0^{\pi/2}\sin^{2n}x(1-\cos^2x)dx\\ &=I_n-\int_0^{\pi/2}\sin^{2n}x\cos^2x~dx\\ &=I_n-\left.\frac1{2n+1}\sin^{2n+1}x\cos x\right|_0^{\pi/2}-\frac1{2n+1}\int_0^{\pi/2}\sin^{2n+1}x\sin x~dx\\ &=I_n-\frac1{2n+1}I_{n+1}\end{align}$$ Solve the recurrent relation and obtain $$I_{n+1}=\frac{2n+1}{2n+2}I_n=\frac{(2n+1)!!}{(2n+2)!!}\frac\pi2$$