第5章 5.3

(1)

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

(2)

$\int_1^{\sqrt{2}} \frac{dx}{x^2\sqrt{4-x^2}}$

$x=2\sin t$ とおくと $dx=2\cos t\,dt$ $x:1\to\sqrt{2}$ $t:\dfrac{\pi}{6}\to\dfrac{\pi}{4}$ これから

$=\int_{\pi/6}^{\pi/4}\frac{2\cos t}{4\sin^2 t\sqrt{4-4\sin^2 t}}\,dt=\int_{\pi/6}^{\pi/4}\frac{2\cos t}{4\sin^2 t\cdot 2\cos t}\,dt=\int_{\pi/6}^{\pi/4}\frac{1}{4\sin^2 t}\,dt=\frac{1}{4}\left[-\frac{1}{\tan t}\right]_{\pi/6}^{\pi/4}=\frac{\sqrt{3}-1}{4}$

(3)

$\int_0^2 \sqrt{4-x^2}\,dx=\left[\frac{1}{2}\left\{x\sqrt{4-x^2}+4\sin^{-1}\frac{x}{2}\right\}\right]_0^2=\frac{1}{2}\left(4\cdot\frac{\pi}{2}\right)=\pi$

(4)

$\int_0^a x^2\sqrt{a^2-x^2}\,dx$

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

$=\int_0^{\pi/2} a^2\sin^2 t\sqrt{a^2-a^2\sin^2 t}\cdot a\cos t\,dt=\int_0^{\pi/2} a^4\sin^2 t\cdot\cos^2 t\,dt=\int_0^{\pi/2} a^4(1-\cos^2 t)\cdot\cos^2 t\,dt$

$=\int_0^{\pi/2} a^4(\cos^2 t-\cos^4 t)\,dt=a^4\left(\frac{1}{2}\cdot\frac{\pi}{2}-\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}\right)=a^4\left(\frac{\pi}{4}-\frac{3}{16}\pi\right)=\frac{\pi a^4}{16}$