삼각치환

(a) We'll use the trig identity $\cos2\theta=2\cos^2\theta-1$. It implies that

$$\begin{aligned} \cos^2\theta=\frac{\cos2\theta+1}{2} \implies \cos^4\theta &=\frac{1}{4}\big[\cos^22\theta+2\cos2\theta+1\big] =\frac{1}{4}\Big[\frac{\cos4\theta+1}{2}+2\cos2\theta+1\Big]\\ &=\frac{\cos4\theta}{8}+\frac{\cos2\theta}{2}+\frac{3}{8} \\ \text{So,} \\ \int_0^{\pi/4}\cos^4\theta\ \mathrm{d}\theta &=\int_0^{\pi/4}\Big(\frac{\cos4\theta}{8}+\frac{\cos2\theta}{2}+\frac{3}{8}\Big) \ \mathrm{d}\theta \\ &=\left[\frac{\sin4\theta}{32}+\frac{\sin2\theta}{4}+\frac{3}{8}\theta\right]_0^{\pi/4}\\ &= \frac{1}{4}+\frac{3}{8}\cdot \frac{\pi}{4}\\ &=\frac{8+3\pi}{32} \end{aligned}$$

as required.

(b) We'll use the trig substitution $x=\tan\theta$, $\mathrm{d}x=\sec^2\theta\ \mathrm{d}\theta$. Note that when $\theta=\pm\frac{\pi}{4}$, we have $x=\pm 1$. Also note that dividing the trig identity $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ gives the trig identity $\tan^2\theta+1=\sec^2\theta$. So

$$\begin{aligned} \int_{-1}^1\frac{\mathrm{d}x}{{(x^2+1)}^3} &=2\int_0^1\frac{\mathrm{d}x}{{(x^2+1)}^3}&\text{(even integrand)}\\ &=2\int_0^{\pi/4}\frac{\sec^2\theta\ \mathrm{d}\theta}{{(\tan^2\theta+1)}^3}\\ &=2\int_0^{\pi/4}\frac{\sec^2\theta\ \mathrm{d}\theta}{{(\sec^2\theta)}^3}\\ &=2\int_0^{\pi/4}\cos^4\theta\ \mathrm{d}\theta\\ &=\frac{8+3\pi}{16} \end{aligned}$$

by part (a).