高専の数学解説

第4章 4.12

(1)

tan⁡−1x\tan^{-1}x

∫tan⁡−1x dx=xtan⁡−1x−∫x1+x2dx=xtan⁡−1x−∫2x1+x2⋅12dx\int \tan^{-1}x\,dx = x\tan^{-1}x - \int \frac{x}{1+x^2}dx = x\tan^{-1}x - \int \frac{2x}{1+x^2}\cdot\frac{1}{2}dx

=xtan⁡−1x−12log⁡(1+x2)+C= x\tan^{-1}x - \frac{1}{2}\log(1+x^2) + C

（C は積分定数）

(2)

xtan⁡−1xx\tan^{-1}x

∫xtan⁡−1x dx=12x2tan⁡−1x−12∫x2x2+1dx=12x2tan⁡−1x−12∫(1−1x2+1)dx\int x\tan^{-1}x\,dx = \frac{1}{2}x^2\tan^{-1}x - \frac{1}{2}\int \frac{x^2}{x^2+1}dx = \frac{1}{2}x^2\tan^{-1}x - \frac{1}{2}\int\left(1-\frac{1}{x^2+1}\right)dx

=12x2tan⁡−1x−12(x−tan⁡−1x)+C=12(x2tan⁡−1x−x+tan⁡−1x)+C= \frac{1}{2}x^2\tan^{-1}x - \frac{1}{2}(x-\tan^{-1}x) + C = \frac{1}{2}(x^2\tan^{-1}x - x + \tan^{-1}x) + C （C は積分定数）

(3)

sin⁡−1x\sin^{-1}x

∫sin⁡−1x dx=xsin⁡−1x−∫x1−x2dx=xsin⁡−1x−1−x2+C\int \sin^{-1}x\,dx = x\sin^{-1}x - \int \frac{x}{\sqrt{1-x^2}}dx = x\sin^{-1}x - \sqrt{1-x^2} + C

（C は積分定数）

(4)

xsin⁡−1xx\sin^{-1}x

∫xsin⁡−1x dx=12x2sin⁡−1x−12∫x21−x2dxx=sin⁡t とおくとdx=cos⁡t dt これから\int x\sin^{-1}x\,dx = \frac{1}{2}x^2\sin^{-1}x - \frac{1}{2}\int \frac{x^2}{\sqrt{1-x^2}}dx \qquad x=\sin t \text{ とおくと} \qquad dx=\cos t\,dt \text{ これから}

=12x2sin⁡−1x−12∫sin⁡2tcos⁡t⋅cos⁡t dt=12x2sin⁡−1x−12∫sin⁡2t dt=12x2sin⁡−1x−12∫1−cos⁡2t2dt= \frac{1}{2}x^2\sin^{-1}x - \frac{1}{2}\int \frac{\sin^2 t}{\cos t}\cdot\cos t\,dt = \frac{1}{2}x^2\sin^{-1}x - \frac{1}{2}\int \sin^2 t\,dt = \frac{1}{2}x^2\sin^{-1}x - \frac{1}{2}\int \frac{1-\cos 2t}{2}dt

=12x2sin⁡−1x−12{12t−14sin⁡2t}+C元に戻して= \frac{1}{2}x^2\sin^{-1}x - \frac{1}{2}\left\{\frac{1}{2}t - \frac{1}{4}\sin 2t\right\} + C \qquad \text{元に戻して}

=12x2sin⁡−1x−12{12sin⁡−1x−14sin⁡(2sin⁡−1x)}+C=14(2x2−1)sin⁡−1x+18sin⁡(sin⁡−1x+sin⁡−1x)+C= \frac{1}{2}x^2\sin^{-1}x - \frac{1}{2}\left\{\frac{1}{2}\sin^{-1}x - \frac{1}{4}\sin(2\sin^{-1}x)\right\} + C = \frac{1}{4}(2x^2-1)\sin^{-1}x + \frac{1}{8}\sin(\sin^{-1}x+\sin^{-1}x) + C

=14(2x2−1)sin⁡−1x+14sin⁡(sin⁡−1x)⋅cos⁡(sin⁡−1x)+C=14{(2x2−1)sin⁡−1x+x1−x2}+C= \frac{1}{4}(2x^2-1)\sin^{-1}x + \frac{1}{4}\sin(\sin^{-1}x)\cdot\cos(\sin^{-1}x) + C = \frac{1}{4}\left\{(2x^2-1)\sin^{-1}x + x\sqrt{1-x^2}\right\} + C

（C は積分定数）

解説: r31bn1z