第5章 5.1

(1)

$\int_0^1 \frac{dx}{\sqrt{4-x^2}} = \left[\sin^{-1}\frac{x}{2}\right]_0^1 = \frac{\pi}{6}$

(2)

$\int_{1/2}^{\sqrt{3}/2} \frac{dx}{\sqrt{1-x^2}} = \left[\sin^{-1}x\right]_{1/2}^{\sqrt{3}/2} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$

(3)

$\int_{-2}^{2} \frac{dx}{x^{2}+4} = \left[\frac{1}{2}\tan^{-1}\frac{x}{2}\right]_{-2}^{2} = \frac{1}{2}\left(\frac{\pi}{4}+\frac{\pi}{4}\right) = \frac{\pi}{4}$

(4)

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

(5)

$\int_0^{\pi/2} \frac{\cos x}{1+\sin^2 x}dx$

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

$\int_0^1 \frac{1}{1+t^2}dt = \left[\tan^{-1}t\right]_0^1 = \frac{\pi}{4}$

(6)

$\int_2^3 \frac{dx}{x^2-6x+10} = \int_2^3 \frac{dx}{(x-3)^2+1} = \left[\tan^{-1}(x-3)\right]_2^3 = 0 - \left(-\frac{\pi}{4}\right) = \frac{\pi}{4}$

(7)

$\int_4^5 \frac{dx}{(x-1)(x-3)} = \int_4^5 \left(\frac{-1}{x-1}+\frac{1}{x-3}\right)\cdot\frac{1}{2}dx = \frac{1}{2}\left[\log\frac{x-3}{x-1}\right]_4^5 = \frac{1}{2}(\log 3 - \log 2) = \frac{1}{2}\log\frac{3}{2}$

(8)

$\int_0^1 \frac{e^x}{1+e^{2x}}dx$

$e^x = t$ とおくと $e^x dx = dt$ $x:0\to1$ $t:1\to e$ これから

$\int_1^e \frac{1}{1+t^2}dt = \left[\tan^{-1}t\right]_1^e = \tan^{-1}e - \frac{\pi}{4}$