3.8 解答

$\begin{displaymath}\begin{array}{llll} x&0 &\Rightarrow & \frac{\pi}{2} \hline t&1 &\Rightarrow & 0 \end{array}\end{displaymath}$

$\displaystyle \int_{0}^{\frac{\pi}{2}}{\cos^{4}{x}\sin{x}} dx$	$\displaystyle =$	$\displaystyle \int_{1}^{0}{t^{4}(-dt)}$
	$\displaystyle =$	$\displaystyle \int_{0}^{1}t^{4} dt = \frac{t^{5}}{5}\mid_{0}^{1} = \frac{1}{5}$

(b) $t = \sqrt{1 + \cos{x}}$ とおくと $t^2 = 1 + \cos{x}$ より $2tdt = -\sin{x}dx$ ．このとき

$\begin{displaymath}\begin{array}{llll} x&0 &\Rightarrow & \frac{\pi}{2} \hline t&\sqrt{2} &\Rightarrow & 1 \end{array}\end{displaymath}$

$\displaystyle \int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sqrt{1 + \cos{x}}}} dx$	$\displaystyle =$	$\displaystyle \int_{\sqrt{2}}^{1}{\frac{-2tdt}{t}}$
	$\displaystyle =$	$\displaystyle -2\int_{0}^{\frac{\pi}{2}} dt = -2t\mid_{\sqrt{2}}^{1} = -2(1 - \sqrt{2}) = 2(\sqrt{2} - 1)$

$\displaystyle \int_{0}^{\frac{\pi}{2}}{\sin^{4}{x}} dx$

$\displaystyle =$

$\displaystyle \frac{3!!}{4!!}\frac{\pi}{2} = \frac{3\pi}{16}$

$\displaystyle \int{f(x)g'(x)} dx = f(x)g(x) - \int{g(x)f'(x)} dx$

$\begin{displaymath}\begin{array}{ll} f(x) = x^2 & g'(x) = \cos{x}\\ f'(x) = 2x & g(x) = \sin{x} \end{array}\end{displaymath}$

$\displaystyle \int_{-1}^{1}{x^{2}\cos{x}} dx$	$\displaystyle =$	$\displaystyle x^{2}\sin{x}\mid_{-1}^{1} - \int{2x\sin{x}} dx$
	$\displaystyle =$	$\displaystyle \sin{1} - \sin{(-1)} - \int{2x\sin{x}} dx = 2\sin{1} - \int{2x\sin{x}} dx$

$\begin{displaymath}\begin{array}{ll} f(x) = 2x & g'(x) = \sin{x}\\ f'(x) = 2 & g(x) = -\cos{x} \end{array}\end{displaymath}$

$\displaystyle \int{2x\sin{x}} dx$	$\displaystyle =$	$\displaystyle -2x\cos{x}\mid_{-1}^{1} + 2\int{\cos{x}} dx$
	$\displaystyle =$	$\displaystyle -2\cos{1} - 2\cos{(-1)} + 2\sin{x}\mid_{-1}^{1}$
	$\displaystyle =$	$\displaystyle -2\cos{1} - 2\cos{1} + 2\sin{1} - 2\sin{(-1)} = -4\cos{1} + 4\sin{1}$

$\displaystyle \int_{-1}^{1}{x^{2}\cos{x}} dx$	$\displaystyle =$	$\displaystyle 2\sin{1} -[-4\cos{1} + 4\sin{1}]$
	$\displaystyle =$	$\displaystyle 4\cos{1} - 2\sin{1}$

$\displaystyle \int_{0}^{\pi}\cos{nx} dx$

$\displaystyle =$

$\displaystyle \frac{1}{n}\sin{nx}\mid_{0}^{\pi} = 0$

2. $\displaystyle{I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{n}{x}dx = \int_{0}^{\frac{\pi}{2}}\sin^{n-1}{x} \sin{x}dx = \frac{n-1}{n} I_{n-2}}$

$\displaystyle{J_{n} = \int_{0}^{\frac{\pi}{2}}\cos^{n}{x}dx = \int_{0}^{\frac{\pi}{2}}\cos^{n-1}{x} \cos{x}dx = \frac{n-1}{n} J_{n-2}}$