第9章 9.7

(1)

$\int_0^{\frac{\pi}{2}} \frac{\sin^3 x}{1+\cos x}dx = \int_0^{\frac{\pi}{2}} \frac{\sin x(1-\cos^2 x)}{1+\cos x}dx = \int_0^{\frac{\pi}{2}} \frac{\sin x(1-\cos x)(1+\cos x)}{1+\cos x}dx = \int_0^{\frac{\pi}{2}} \sin x(1-\cos x)dx$

$\cos x = t$ とおくと $-\sin x\, dx = dt$ $x: 0 \to \dfrac{\pi}{2}$ $t: 1 \to 0$ これから

$= \int_1^0 -(1-t)dt = \left[-t+\frac{1}{2}t^2\right]_1^0 = -\left(-1+\frac{1}{2}\right) = \frac{1}{2}$

(2)

$\int_0^{\frac{\pi}{3}} \frac{\sin^3 x}{1+2\cos x}dx = \int_0^{\frac{\pi}{3}} \frac{\sin x(1-\cos^2 x)}{1+2\cos x}dx$

$\cos x = t$ とおくと $-\sin x\, dx = dt$ $x: 0 \to \dfrac{\pi}{3}$ $t: 1 \to \dfrac{1}{2}$ これから

$= \int_1^{\frac{1}{2}} -\frac{1-t^2}{1+2t}dt = \int_{\frac{1}{2}}^1 \frac{1-t^2}{1+2t}dt = \int_{\frac{1}{2}}^1 \left(-\frac{1}{2}t+\frac{1}{4}+\frac{\frac{3}{4}}{2t+1}\right)dt = \left[-\frac{1}{4}t^2+\frac{1}{4}t+\frac{3}{8}\log|2t+1|\right]_{\frac{1}{2}}^1$

$= -\frac{1}{4}+\frac{1}{4}+\frac{3}{8}\log 3+\frac{1}{16}-\frac{1}{8}-\frac{3}{8}\log 2 = -\frac{1}{16}+\frac{3}{8}(\log 3-\log 2)$