高専の数学解説

第8章 8.12

(1)

∫x1−x dx=x⋅(−23)(1−x)32+∫23(1−x)32dx=−23x(1−x)32−23⋅25(1−x)52+C\int x\sqrt{1-x}\,dx = x\cdot\left(-\frac{2}{3}\right)(1-x)^{\frac{3}{2}} + \int \frac{2}{3}(1-x)^{\frac{3}{2}}dx = -\frac{2}{3}x(1-x)^{\frac{3}{2}} - \frac{2}{3}\cdot\frac{2}{5}(1-x)^{\frac{5}{2}} + C

=−215{5x(1−x)+2(1−x)2}1−x+C=−215{5x−5x2+2−4x+2x2}1−x+C= -\frac{2}{15}\{5x(1-x)+2(1-x)^2\}\sqrt{1-x} + C = -\frac{2}{15}\{5x-5x^2+2-4x+2x^2\}\sqrt{1-x}+C

=−215(−3x2+x+2)1−x+C=215(3x2−x−2)1−x+C(Cは積分定数)= -\frac{2}{15}(-3x^2+x+2)\sqrt{1-x}+C = \frac{2}{15}(3x^2-x-2)\sqrt{1-x}+C \qquad (C\text{は積分定数})

(2)

∫(log⁡x)2dx=x(log⁡x)2−∫x⋅2log⁡x⋅1xdx=x(log⁡x)2−2(xlog⁡x−x)+C=x(log⁡x)2−2x⋅log⁡x+2x+C\int (\log x)^2 dx = x(\log x)^2 - \int x\cdot 2\log x\cdot\frac{1}{x}dx = x(\log x)^2 - 2(x\log x - x) + C = x(\log x)^2 - 2x\cdot\log x + 2x + C

(Cは積分定数)(C\text{は積分定数})

(3)

∫x⋅tan⁡2x dx\int x\cdot\tan^2 x\,dx

∫tan⁡2x dx=∫sin⁡2xcos⁡2xdx=∫1−cos⁡2xcos⁡2xdx=∫1cos⁡2xdx−∫dx=tan⁡x−x+C\int \tan^2 x\,dx = \int \frac{\sin^2 x}{\cos^2 x}dx = \int \frac{1-\cos^2 x}{\cos^2 x}dx = \int \frac{1}{\cos^2 x}dx - \int dx = \tan x - x + C

これから

∫x⋅tan⁡2x dx=x(tan⁡x−x)−∫(tan⁡x−x)dx=x⋅tan⁡x−x2−∫tan⁡x dx+∫x dx\int x\cdot\tan^2 x\,dx = x(\tan x - x) - \int (\tan x - x)dx = x\cdot\tan x - x^2 - \int \tan x\,dx + \int x\,dx

=x⋅tan⁡x−x2+∫sin⁡xcos⁡xdx+12x2=x⋅tan⁡x−x2+log⁡∣cos⁡x∣+12x2+C= x\cdot\tan x - x^2 + \int \frac{\sin x}{\cos x}dx + \frac{1}{2}x^2 = x\cdot\tan x - x^2 + \log|\cos x| + \frac{1}{2}x^2 + C

=−12x2+x⋅tan⁡x+log⁡∣cos⁡x∣+C(Cは積分定数)= -\frac{1}{2}x^2 + x\cdot\tan x + \log|\cos x| + C \qquad (C\text{は積分定数})

(4)

∫x⋅cos⁡x⋅sin⁡x dx=∫x⋅12sin⁡2x dx=12⋅12x(−cos⁡2x)+∫14cos⁡2x dx=−14x⋅cos⁡2x+14⋅12sin⁡2x+C\int x\cdot\cos x\cdot\sin x\,dx = \int x\cdot\frac{1}{2}\sin 2x\,dx = \frac{1}{2}\cdot\frac{1}{2}x(-\cos 2x) + \int \frac{1}{4}\cos 2x\,dx = -\frac{1}{4}x\cdot\cos 2x + \frac{1}{4}\cdot\frac{1}{2}\sin 2x + C

=18(−2xcos⁡2x+sin⁡2x)+C(Cは積分定数)= \frac{1}{8}(-2x\cos 2x + \sin 2x) + C \qquad (C\text{は積分定数})

(5)

∫tan⁡4xcos⁡2xdxtan⁡x=t とおくと 1cos⁡2xdx=dt となるので\int \frac{\tan^4 x}{\cos^2 x}dx \qquad \tan x = t\text{ とおくと }\frac{1}{\cos^2 x}dx = dt\text{ となるので}

=∫tan⁡4x⋅1cos⁡2xdx=∫t4dt=15t5+C=15tan⁡5x+C(Cは積分定数)= \int \tan^4 x\cdot\frac{1}{\cos^2 x}dx = \int t^4 dt = \frac{1}{5}t^5 + C = \frac{1}{5}\tan^5 x + C \qquad (C\text{は積分定数})

(6)

∫1sin⁡x⋅cos⁡x dx\int \frac{1}{\sin x \cdot \cos x}\, dx

sin⁡x=t\sin x = t とおくと cos⁡x dx=dt\cos x\, dx = dt となるので

=∫cos⁡xsin⁡x⋅cos⁡2x dx=∫1sin⁡x(1−sin⁡2x)cos⁡x dx=∫1t(1−t2) dt=∫(1t+t1−t2)dt= \int \frac{\cos x}{\sin x \cdot \cos^2 x}\, dx = \int \frac{1}{\sin x(1-\sin^2 x)}\cos x\, dx = \int \frac{1}{t(1-t^2)}\, dt = \int \left(\frac{1}{t} + \frac{t}{1-t^2}\right) dt

=∫1t dt+12∫2t1−t2 dt=log⁡∣t∣−12log⁡∣1−t2∣+C= \int \frac{1}{t}\, dt + \frac{1}{2}\int \frac{2t}{1-t^2}\, dt = \log|t| - \frac{1}{2}\log|1-t^2| + C

元に戻して

=log⁡∣sin⁡x∣−12log⁡∣1−sin⁡2x∣+C=log⁡∣sin⁡x∣−log⁡∣cos⁡x∣+C=log⁡∣sin⁡xcos⁡x∣+C=log⁡∣tan⁡x∣+C= \log|\sin x| - \frac{1}{2}\log|1-\sin^2 x| + C = \log|\sin x| - \log|\cos x| + C = \log\left|\frac{\sin x}{\cos x}\right| + C = \log|\tan x| + C

（Cは積分定数）

(7)

∫3xx2−x−2 dx=∫3x(x−2)(x+1) dx=∫(2x−2+1x+1)dx=2log⁡∣x−2∣+log⁡∣x+1∣+C\int \frac{3x}{x^2 - x - 2}\, dx = \int \frac{3x}{(x-2)(x+1)}\, dx = \int \left(\frac{2}{x-2} + \frac{1}{x+1}\right) dx = 2\log|x-2| + \log|x+1| + C

=log⁡(x−2)2∣x+1∣+C= \log (x-2)^2|x+1| + C

（Cは積分定数）

(8)

∫ex{x−log⁡(ex+2)} dx\int e^x \{x - \log(e^x+2)\}\, dx

ex=te^x = t とおくと x=log⁡tx = \log t、exdx=dte^x dx = dt から

=∫{log⁡t−log⁡(t+2)} dt=t⋅log⁡t−t−{(t+2)⋅log⁡(t+2)−(t−2)}+C= \int \{\log t - \log(t+2)\}\, dt = t \cdot \log t - t - \{(t+2)\cdot \log(t+2) - (t-2)\} + C

=t⋅log⁡t−(t+2)⋅log⁡(t+2)+C=ex⋅x−(ex+2)⋅log⁡(ex+2)+C=xex−(ex+2)log⁡(ex+2)+C= t \cdot \log t - (t+2)\cdot \log(t+2) + C = e^x \cdot x - (e^x+2)\cdot \log(e^x+2) + C = xe^x - (e^x+2)\log(e^x+2) + C

（Cは積分定数）

解説: r31bn1z