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3.2 解答

3.2

1.

(a) $ t = 2 - x$ とおくと $ dt = \frac{dt}{dx}dx = -dx$ より，与えられた積分はtの関数とdtで，すでに学んだ$ e^{t}$の$ t$についての積分として表せる．

$\displaystyle \int e^{2-x} dx = - \int e^{t} dt = - e^{t} + c \underbrace{=}_{t = 2-xを代入} {e^{2 - x}} + c $

(b)

$ t = 1 - x$ とおくと $ dt = -dx$ より， $ \sec^{2}{t}$の$ t$についての積分として表せる．

$\displaystyle \int \sec^{2}{(1 - x)}dx = - \int \sec^{2}{t} dt = -\tan (1 - x) + c $

(c)

$ t = 1 - x^2$ とおくと $ dt = -2xdx$ より

$\displaystyle \int \frac{x}{\sqrt{1 - x^2}} dx = -\frac{1}{2} \int \frac{dt}{\sqrt{t}} = -\frac{1}{2} \int t^{-\frac{1}{2}} dt = -{\sqrt{1 - {x^2}}} + c$

(d)

$ t = \cos{x}$ とおくと $ dt = -\sin{x}dx$ より

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

(e)

$ t = \frac{1}{x}$ とおくと $ dt = - \frac{1}{x^2} dx$ より

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

(f)

$ t = 3\tan{\theta} + 1$ とおくと $ dt = 3\sec^{2}{\theta} d \theta$より

$\displaystyle \int \frac{sec^{2}{\theta}}{3\tan{\theta} + 1} d \theta = \frac{1... ...t}} = \frac{2}{3}\sqrt{t} + c = \frac{2 {\sqrt{1 + 3 \tan (\theta)}}}{3} + c$

(g)


$\displaystyle \int \frac{1 + \cos{2x}}{\sin^{2}{x}}dx$ $\displaystyle =$ $\displaystyle \int \frac{\cos^{2}{x}}{\sin^{2}{x}}dx$  
  $\displaystyle =$ $\displaystyle 2\int \frac{1 - \sin^{2}{x}}{\sin^{2}{x}}dx = 2\int({\rm cosec}^{2}{x} - 1)dx$  
  $\displaystyle =$ $\displaystyle 2(-\cot{x} - x) + c$  

(h)

$ t = \log{x}$ とおくと $ dt = \frac{1}{x}dx$ より

$\displaystyle \int \frac{\log{x}}{x}dx = \int t dt = \frac{1}{t^{2}} + c = \frac{1}{2}(\log (x))^2 + c $

(i)

$ t = e^{x}$ とおくと $ dt = e^{x} dx$ より

$\displaystyle \int \frac{e^{x}}{1 + e^{2x}}dx = \int \frac{1}{1 + t^{2}} dt = \tan^{-1} ({e^x}) + c$

(j)

$ t = x^2$ とおくと $ dt = 2x dx$ より

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