We are going to prove Wallis’s formula for odd powers with \(n\ge3\):
\[\int_0^{\pi/2}\cos^nx\mbox{ d}x=\frac{2}{3}\cdot\frac{4}{5}\cdot\cdots\cdot\frac{n-1}{n}\]
We are going to use the following formula (problem 80):
\[\int\cos^n x\mbox{ d}x=\frac{\cos^{n-1}x\sin x}{n}+\frac{n-1}{n}\int\cos^{n-2}x\mbox{ d}x\]
We are going to do this in two steps (this is known as proof by induction):
Prove that the formula is true for \(n=3\)
Prove that if the formula is true for \(n-2\), it is also true for \(n\).
\[\begin{aligned} \int_0^{\pi/2}\cos^3 x\mbox{ d}x &=\left.\frac{\cos^2x\sin x}{3}\right|_0^{\pi/2}+\frac{2}{3}\int_0^{\pi/2}\cos x\mbox{ d}x\\ &=\left.\frac{\cos^2x\sin x}{3}+\frac{2}{3}\sin x\right|_0^{\pi/2}\\ &=\left(\frac{\cos^2\frac{\pi}{2}\sin\frac{\pi}{2}}{3}+\frac{2}{3}\sin\frac{\pi}{2}\right)-\left(\frac{\cos^2 0\sin0}{3}+\frac{2}{3}\sin 0\right)\\ &=\left(\frac{0\cdot1}{3}+\frac{2}{3}\cdot1\right)-\left(\frac{1\cdot0}{3}+\frac{2}{3}\cdot0\right)\\ &=\frac{2}{3}-0\\ &=\frac{2}{3}\\\end{aligned}\]
Assuming the formula is true for \(n-2\), that means
\[\int_0^{\pi/2}\cos^{n-2}x\mbox{ d}x=\frac{2}{3}\cdot\frac{4}{5}\cdot\cdots\cdot\frac{n-3}{n-2}.\]
Now we want to find \(\int_0^{\pi/2}\cos^{n-2}x\mbox{ d}x\). Using the formula from problem 80, we get:
\[\begin{aligned} \int_0^{\pi/2}\cos^n x\mbox{ d}x &=\left.\frac{\cos^{n-1}x\sin x}{n}\right|_0^{\pi/2}+\frac{n-1}{n}\int_0^{\pi/2}\cos^{n-2}x\mbox{ d}x\\ &=\left(\frac{\cos^{n-1}\frac{\pi}{2}\sin \frac{\pi}{2}}{n}\right)-\left(\frac{\cos^{n-1}0\sin0}{n}\right)+\frac{n-1}{n}\frac{2}{3}\cdot\frac{4}{5}\cdot\cdots\cdot\frac{n-3}{n-2}\\ &=\left(\frac{0\cdot1}{n}\right)-\left(\frac{1\cdot0}{n}\right)+\frac{2}{3}\cdot\frac{4}{5}\cdot\cdots\cdot\frac{n-3}{n-2}\frac{n-1}{n}\\ &=\frac{2}{3}\cdot\frac{4}{5}\cdot\cdots\cdot\frac{n-3}{n-2}\frac{n-1}{n}\end{aligned}\]
This completes the proof because it means the formula is true for \(n=3\) by the first part. By the second part it is then true for \(n=5\), since it is true for \(n=5\), it is true for \(n=7\), and so on.
\(\square\)