第8章 8.8

(1)

$\int_0^a\int_{\frac{x^2}{a}}^{x} \frac{x}{x^2+y^2}\,dy\,dx = \int_0^a\left[x\cdot\frac{1}{x}\tan^{-1}\frac{y}{x}\right]_{\frac{x^2}{a}}^{x}dx = \int_0^a \tan^{-1}1 - \tan^{-1}\frac{x}{a}\,dx = \int_0^a \frac{\pi}{4}-\tan^{-1}\frac{x}{a}\,dx$

$=\left[\frac{\pi}{4}x\right]_0^a - \left[x\tan^{-1}\frac{x}{a}\right]_0^a + \int_0^a \frac{ax}{a^2+x^2}\,dx = \frac{\pi}{4}a-\frac{\pi}{4}a+\left[\frac{a}{2}\log(a^2+x^2)\right]_0^a$

$=\frac{a}{2}(\log 2a^2 - \log a^2) = \frac{a}{2} \log 2$

(2)

$\int_0^1\int_y^1 x^2\cos(x^2-xy)\,dx\,dy = \int_0^1\int_0^x x^2\cos(x^2-xy)\,dy\,dx = \int_0^1\left[-x\sin(x^2-xy)\right]_0^x dx = \int_0^1 x\sin x^2\,dx$

$x^2 = t$ とおくと $2xdx = dt$ $x:0\to1$ $t:0\to1$ これから

$=\int_0^1 \frac{1}{2}\sin t\,dt = \left[-\frac{1}{2}\cos t\right]_0^1 = \frac{1}{2}(1-\cos 1)$