부분적분법

(a) The “parts" in the integrand are powers of sine. Looking at the right hand side of the reduction formula, we see that it looks a little like the derivative of $\sin^{n-1}x$, although not exactly. So, let's integrate by parts with $u=\sin^{n-1}x$ and $\mathrm{d}v=\sin x\,\mathrm{d}x$, so that $\mathrm{d}u=(n-1)\sin^{n-2}x\cos x$ and $v=-\cos x$.

$$\begin{aligned} \int\sin^nx\,\mathrm{d}x &=\underbrace{-\sin^{n-1}x\ \cos x}_{uv}+\underbrace{(n-1)\int \cos^2x\ \sin^{n-2}x\ \mathrm{d}x}_{-\int v\mathrm{d}u} \\ \text{Using the identity $\sin^2 x + \cos^2 x =1 $,} \\ &=-\sin^{n-1}x\ \cos x+(n-1)\int (1-\sin^2 x)\sin^{n-2}x\ \mathrm{d}x\\ &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \mathrm{d}x -(n-1)\int\sin^{n}x\ \mathrm{d}x \end{aligned}$$

Moving the last term on the right hand side to the left hand side gives

$$\begin{aligned} n\int\sin^nx\,\mathrm{d}x &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \mathrm{d}x \end{aligned}$$

Dividing across by $n$ gives the desired reduction formula.

(b) By the reduction formula of part (a), if $n \ge 2$,

$$\begin{aligned} \int_0^{\pi/2}\sin^n(x)\,\mathrm{d}x= \frac{n-1}{n}\int_0^{\pi/2}\sin^{n-2}(x)\,\mathrm{d}x \end{aligned}$$

since $\sin 0=\cos\frac{\pi}{2}=0$. Applying this reduction formula, with $n=8,6,4,2$:

$$\begin{aligned} \int_0^{\pi/2}\sin^8(x)\,\mathrm{d}x &=\frac{7}{8}\int_0^{\pi/2}\sin^6(x)\,\mathrm{d}x =\frac{7}{8}\cdot\frac{5}{6}\int_0^{\pi/2}\sin^4(x)\,\mathrm{d}x =\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\int_0^{\pi/2}\sin^2(x)\,\mathrm{d}x \\ &=\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\int_0^{\pi/2}\,\mathrm{d}x =\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} =\frac{35}{256}\pi \end{aligned}$$

Using a calculator, we see this is approximately $0.4295$.