Reduction formula

[TsAmE]

Use \(\displaystyle \int_{0}^{\frac{\pi}{2}}sin^{n}xdx = \frac{n - 1}{n} \int_{0}^{\frac{\pi}{2}}sin^{n-2}xdx\)

to show that, for odd powers of sine,

\(\displaystyle \int_{0}^{\frac{\pi}{2}}sin^{2n + 1}xdx = \frac{2 \cdot 4 \cdot 6 \cdot \cdot \cdot \cdot \cdot 2n }{3\cdot5\cdot7\cdot\cdot\cdot\cdot\cdot(2n + 1)}\)

Attempt:

I subbed in 2n + 1 into n in the original reduction formula and got:

\(\displaystyle \int_{0}^{\frac{\pi}{2}}sin^{2n+1}xdx = \frac{2n}{2n +1} \int_{0}^{\frac{\pi}{2}}sin^{2n-1}xdx\)

but I cant see what more I can do

 

[galactus]

I will show for \(\displaystyle \int_{0}^{\frac{\pi}{2}}sin^{n}(x)dx=\frac{2\cdot 4\cdot 6\cdot\cdot\cdot (n-1)}{3\cdot 5\cdot 7\cdot\cdot\cdot n}\)
and you can try and adapt it to \(\displaystyle sin^{2n+1}(x)\). Okey-doke.

\(\displaystyle \int_{0}^{\frac{\pi}{2}}sin^{n}(x)dx=\frac{-1}{n}sin^{n-1}(x)cos(x)|_{0}^{\frac{\pi}{2}}+\frac{n-1}{n}\int_{0}^{\frac{\pi}{2}}sin^{n-2}(x)dx=\frac{n-1}{n}\int_{0}^{\frac{\pi}{2}}sin^{n-2}(x)dx\)

By repeating the application of the above formula:

\(\displaystyle \int_{0}^{\frac{\pi}{2}}sin^{n}(x)dx=\left(\frac{n-1}{n}\right)\left(\frac{n-3}{n-2}\right)\int_{0}^{\frac{\pi}{2}}sin^{n-4}(x)dx\)

For n odd:

\(\displaystyle \left(\frac{n-1}{n}\right)\left(\frac{n-3}{n-2}\right)\left(\frac{n-5}{n-4}\right)\cdot\cdot\cdot \frac{2}{3}\underbrace{\int_{0}^{\frac{\pi}{2}}sin(x)dx}_{\text{this equals 1}}\)

\(\displaystyle \frac{2\cdot 4\cdot 6\cdot\cdot\cdot (n-1)}{3\cdot 5\cdot 7\cdot \cdot \cdot n}\)

 

[TsAmE]

\(\displaystyle \int_{0}^{\frac{\pi}{2}}sin^{n}(x)dx=\left(\frac{n-1}{n}\right)\left(\frac{n-3}{n-2}\right)\int_{0}^{\frac{\pi}{2}}sin^{n-4}(x)dx\)

Where did the (n - 3 / n -2) and sin^(n-4)x come from?