Evaluate the integral of sin^n(x) from 0 to pi/2 for a non-negative integer n.

Step-by-step solution

1. 1

   Split off one sine and integrate by parts. Write $\sin^{n}x=\sin^{n-1}x\cdot\sin x$ and take

   $u=\sin^{n-1}x,\qquad dv=\sin x\,dx\ \Longrightarrow\ du=(n-1)\sin^{n-2}x\cos x\,dx,\quad v=-\cos x$

   The boundary term $\left[-\sin^{n-1}x\cos x\right]_{0}^{\pi/2}$ vanishes at both ends — $\cos\tfrac\pi2=0$ at the top and $\sin 0=0$ at the bottom — which is exactly why these limits are chosen.

2. 2

   Convert $\cos^{2}$ back to $\sin$. What remains is

   $I_{n}=(n-1)\int_{0}^{\pi/2}\sin^{n-2}x\cos^{2}x\,dx=(n-1)\int_{0}^{\pi/2}\sin^{n-2}x\left(1-\sin^{2}x\right)dx$

   so

   $I_{n}=(n-1)I_{n-2}-(n-1)I_{n}$

   The original integral reappears on the right, which is the signal to solve for it algebraically rather than integrate again.

3. 3

   Solve for $I_{n}$ to get the reduction formula. Adding $(n-1)I_{n}$ to both sides gives $nI_{n}=(n-1)I_{n-2}$, hence

   $I_{n}=\frac{n-1}{n}\,I_{n-2}$

   Each application lowers the index by two, so the recursion terminates at $I_{0}$ or $I_{1}$ depending on the parity of $n$.

4. 4

   Compute the two base cases.

   $I_{0}=\int_{0}^{\pi/2}1\,dx=\frac{\pi}{2},\qquad I_{1}=\int_{0}^{\pi/2}\sin x\,dx=\left[-\cos x\right]_{0}^{\pi/2}=1$

   These are the sole source of the $\pi$: it survives only in the even chain.

5. 5

   Unroll the recursion into the Wallis formulas.

   $I_{n}=\frac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2}\cdot\frac{\pi}{2}\ \ (n\text{ even}),\qquad I_{n}=\frac{(n-1)(n-3)\cdots 2}{n(n-2)\cdots 3}\ \ (n\text{ odd})$

   For example $I_{4}=\tfrac34\cdot\tfrac12\cdot\tfrac\pi2=\tfrac{3\pi}{16}=0.58905$ and $I_{5}=\tfrac45\cdot\tfrac23=\tfrac{8}{15}=0.53333$.

6. 6

   Note the single formula that covers both parities. Using the Gamma function,

   $\int_{0}^{\pi/2}\sin^{n}x\,dx=\frac{\sqrt{\pi}\,\Gamma\!\left(\frac{n+1}{2}\right)}{2\,\Gamma\!\left(\frac{n}{2}+1\right)}$

   Numerical integration for $n=0$ through $7$ agrees with this to ten decimal places in every case ✓, and it extends the result to non-integer $n>-1$.