3.2 置換積分法

1.

与えられた積分をテキスト99ページの公式の形に変形する．

(a) $t = 2x$，$dt = 2dx$より，

$${\int \sin{2x} \; dx = \int \sin{t} \;\frac{dt}{2} = \frac{1}{2} \int \sin{t} \; dt = \frac{1}{2}(-\cos{t}) + c = -\frac{1}{2}\cos{2x} + c}$$

(b) $t = x^2 + 1$，$dt = 2xdx$より，

$${\int \frac{x}{x^2 + 1} \; dx = \int \frac{dt/2}{t} = \frac{1}{2}\... ...{1}{t}\; dt = \frac{1}{2}\log{\vert t\vert} + c = \frac{1}{2}\log(x^2 + 1) + c}$$

(c) $t = 2x$，$dt = 2dx$より，

$${\int e^{2x} \; dx = \int e^{t} \frac{dt}{2} = \frac{1}{2}\int e^t \; dt = \frac{1}{2}e^t + c = \frac{1}{2}e^{2x} + c}$$

(d) $t = \log{x}$， $dt = \frac{1}{x}dx$より，

$${\int \frac{1}{x\log{x}} \; dx = \int \frac{1}{t} \;dt = \log{\vert t\vert} + c = \log{\vert\log{x}\vert} + c}$$

(e) $t = x^2$，$dt = 2xdx$より，

$${\int xe^{x^2} \; dx = \int e^{t} \;\frac{dt}{2} = \frac{1}{2}\int e^{t} \; dt = \frac{1}{2}e^{t} + c = \frac{1}{2}e^{x^2} + c}$$

(f) $t = \sin{x}$， $dt = \cos{x}dx$より，

$${\int \sin^{2}{x}\cos{x} \; dx = \int t^{2} \;dt = \frac{t^3}{3} + c = \frac{\sin^{3}{x}}{3} + c}$$

(g) $t = 1+x$とおくと，$dt = dx$. また，$x = t - 1$となる．これより，

$$\int x\sqrt{1+x}\;dx = \int (t-1)\sqrt{t}\; dt = \int (t^{3/2} - t^{1/2}) \; dt$$

$$= \frac{2}{5}t^{5/2} - \frac{2}{3}t^{3/2} + c = \frac{2}{5}(1+x)^{5/2} - \frac{2}{3}(1+x)^{3/2} + c$$

(h) $t = \sin{x} + \cos{x}$， $dt = (\cos{x} - \sin{x})dx$より，

$${\int \frac{\cos{x} - \sin{x}}{\sin{x} + \cos{x}} \; dx = \int \frac{1}{t}\; dt = \log{\vert t\vert} + c = \log{\vert\sin{x} + \cos{x}\vert} + c}$$

(i) $t = 1+e^x$， $dt = e^x dx$より，

$${\int \frac{e^x}{1 + e^x} \; dx = \int \frac{1}{t} \;dt = \log{\vert t\vert} + c = \log(1 + e^x) + c}$$