치환적분법

Perhaps shorter ways exist, but the reasoning here is valid.

Problem: Evaluate $\displaystyle\int_{0}^{\pi/4} x\tan (x^2) \ \mathrm{d}x$.

Work: We begin with the substitution $u=x^2$, $\mathrm{d}u = 2x\mathrm{d}x$:

If $u=x^2$, then $\frac{\mathrm{d}u}{\mathrm{d}x} = 2x$, so indeed $\mathrm{d}u=2x\mathrm{d}x$.

$$\begin{aligned} \int_{0}^{\pi/4} x\tan (x^2) \ \mathrm{d}x&= \int_{0}^{\pi/4} \frac{1}{2}\tan(x^2)\cdot 2x\mathrm{d}x&\text{algebra}\\ &=\int_{0}^{\pi^2/16} \frac{1}{2}\tan u\ \mathrm{d}u \\ \text{\text{Every piece is changed from $x$ to $u$: integrand, differential, limits.}} \\ &=\frac{1}{2}\int_{0}^{\pi^2/16} \dfrac{\sin u}{\cos u}\mathrm{d}u & \tan u = \frac{\sin u}{\cos u}\\ \\ \text{Now we use the substitution $v=\cos u$, $\mathrm{d}v=-\sin u \ \mathrm{d}u$:} \\ &=\frac{1}{2}\int_{\cos 0}^{\cos(\pi^2/16)} -\dfrac{1}{v}\mathrm{d}v \\ \text{\text{Every piece is changed from $u$ to $v$: integrand, differential, limits.}} \\ &=-\frac{1}{2}\int_{1}^{\cos(\pi^2/16)} \dfrac{1}{v}\mathrm{d}v& \cos(0)=1\\ &=-\frac{1}{2}\bigg[\log|v|\bigg]_{1}^{\cos(\pi^2/16)}&\text{FTC Part 2}\\ &=-\frac{1}{2}\left(\log\left(\cos(\pi^2/16)\right)-\log(1)\right)\\ &=-\frac{1}{2}\log\left(\cos(\pi^2/16)\right)& \log(1)=0 \end{aligned}$$