적분 전략 고르기

(a) Integrate by parts with $u=x$ and $\mathrm{d}v=\sin x\,\mathrm{d}x$ so that $\mathrm{d}u=\mathrm{d}x$ and $v=-\cos x$.

$$\begin{aligned} \int x\sin x\,\mathrm{d}x&= -x\cos x-\int (-\cos x)\,\mathrm{d}x =-x\cos x+\sin x+C\\ \text{So, }\quad\int_0^{\pi/2} x\sin x\,\mathrm{d}x&=\Big[-x\cos x+\sin x\Big]_0^{\pi/2} =1 \end{aligned}$$

(b) The power of cosine is odd, so we can reserve one cosine for $\mathrm{d}u$ and change the rest into sines. Make the substitution $u=\sin x$, $\mathrm{d}u=\cos x\,\mathrm{d}x$.

$$\begin{aligned} \int_0^{\pi/2} \cos^5 x\,\mathrm{d}x &=\int_0^{\pi/2} \big(1-\sin^2x\big)^2\cos x\,\mathrm{d}x =\int_0^1\big(1-u^2\big)^2\,\mathrm{d}u =\int_0^1\big(1-2u^2+u^4\big)\,\mathrm{d}u\\ &=\left[u-\frac{2}{3}u^3+\frac{1}{5}u^5\right]_0^1 =1-\frac{2}{3}+\frac{1}{5} =\frac{8}{15} \end{aligned}$$