第5章 5.15

(1)

$\int_0^1 \frac{dx}{x^2-x+1} = \int_0^1 \frac{dx}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}} = \left[\frac{2}{\sqrt{3}}\tan^{-1}\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right]_0^1 = \frac{2}{\sqrt{3}}\left(\frac{\pi}{6}+\frac{\pi}{6}\right) = \frac{2}{3\sqrt{3}}\pi$

(2)

$\int_0^1 \frac{x^2}{\sqrt{4-x^2}}dx$

$x=2\sin t$ とおくと $dx=2\cos t\,dt$ $x:0\to1$ $t:0\to\dfrac{\pi}{6}$ これから

$=\int_0^{\frac{\pi}{6}}\frac{4\sin^2 t}{2\cos t}\cdot 2\cos t\,dt=\int_0^{\frac{\pi}{6}}4\sin^2 t\,dt=4\int_0^{\frac{\pi}{6}}\frac{1-\cos 2t}{2}dt=4\left[\frac{1}{2}t-\frac{1}{4}\sin 2t\right]_0^{\frac{\pi}{6}}$

$=4\left\{\frac{\pi}{12}-\frac{1}{4}\left(\frac{\sqrt3}{2}\right)\right\}=\frac{\pi}{3}-\frac{\sqrt3}{2}$

(3)

$\int_0^1 (2+x)\sqrt{1-x^2}\,dx=\left[(2+x)\left\{\frac{1}{2}\left(x\sqrt{1-x^2}+\sin^{-1}x\right)\right\}\right]_0^1-\frac{1}{2}\int_0^1\left(x\sqrt{1-x^2}+\sin^{-1}x\right)dx$

$=\left(\frac{1}{2}\cdot3\cdot\frac{\pi}{2}\right)-\frac{1}{2}\left[-\frac{1}{2}\sqrt{(1-x^2)^3}\cdot\frac{2}{3}+x\sin^{-1}x+\sqrt{1-x^2}\right]_0^1$

$=\frac{3}{4}\pi-\frac{1}{2}\left(\frac{\pi}{2}+\frac{1}{3}-1\right)=\frac{\pi}{2}+\frac{1}{3}$

(4)

$\int_1^{\sqrt3}\tan^{-1}x\,dx=\left[x\tan^{-1}x\right]_1^{\sqrt3}-\int_1^{\sqrt3}\frac{x}{x^2+1}dx=\sqrt3\cdot\frac{\pi}{3}-\frac{\pi}{4}-\left[\frac{1}{2}\log(x^2+1)\right]_1^{\sqrt3}$

$=\frac{4\sqrt3\pi-3\pi}{12}-\frac{1}{2}\log2$

(5)

$\int_0^1 x\tan^{-1}x\,dx=\left[\frac{1}{2}\left(x^2\tan^{-1}x-x+\tan^{-1}x\right)\right]_0^1\quad(4.12(2)\text{から})$

$=\frac{1}{2}\left(\frac{\pi}{4}-1+\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{1}{2}$

(6)

$\int_0^{\frac{1}{2}} \sin^{-1}x \, dx = [x\sin^{-1}x + \sqrt{1-x^2}]_0^{\frac{1}{2}} \qquad (4.12(3)\text{より})$

$= \frac{1}{2}\cdot\frac{\pi}{6} + \frac{\sqrt{3}}{2} - 1 = \frac{\pi}{12} + \frac{\sqrt{3}}{2} - 1$