高専の数学解説

第5章 5.2

(1)

∫0π/2dxsin⁡x+cos⁡x\displaystyle\int_0^{\pi/2}\frac{dx}{\sin x+\cos x} tan⁡x2=t\tan\dfrac{x}{2}=t とおくと dx=21+t2dtdx=\dfrac{2}{1+t^2}dt sin⁡x=2t1+t2\sin x=\dfrac{2t}{1+t^2} cos⁡x=1−t21+t2\cos x=\dfrac{1-t^2}{1+t^2}

x:0→π/2x:0\to\pi/2 t:0→1t:0\to1 これから

=∫0112t1+t2+1−t21+t2⋅21+t2 dt=∫0122t+1−t2 dt=∫0122−(t−1)2 dt=∫01−2(t−1)2−2 dt=\int_0^1\frac{1}{\dfrac{2t}{1+t^2}+\dfrac{1-t^2}{1+t^2}}\cdot\frac{2}{1+t^2}\,dt=\int_0^1\frac{2}{2t+1-t^2}\,dt=\int_0^1\frac{2}{2-(t-1)^2}\,dt=\int_0^1\frac{-2}{(t-1)^2-2}\,dt

=−12[log⁡∣t−1−2t−1+2∣]01=−12(log⁡∣−1∣−log⁡∣−1−2−1+2∣)=12log⁡∣−1−2−1+2∣=-\frac{1}{\sqrt2}\left[\log\left|\frac{t-1-\sqrt2}{t-1+\sqrt2}\right|\right]_0^1=-\frac{1}{\sqrt2}\left(\log|-1|-\log\left|\frac{-1-\sqrt2}{-1+\sqrt2}\right|\right)=\frac{1}{\sqrt2}\log\left|\frac{-1-\sqrt2}{-1+\sqrt2}\right|

=12log⁡∣(−1−2)2(−1+2)(−1−2)∣=12log⁡(1+2)2=22log⁡(1+2)=2log⁡(1+2)=\frac{1}{\sqrt2}\log\left|\frac{(-1-\sqrt2)^2}{(-1+\sqrt2)(-1-\sqrt2)}\right|=\frac{1}{\sqrt2}\log(1+\sqrt2)^2=\frac{2}{\sqrt2}\log(1+\sqrt2)=\sqrt2\log(1+\sqrt2)

(2)

∫0π/2dx2−cos⁡x\displaystyle\int_0^{\pi/2}\frac{dx}{2-\cos x} tan⁡x2=t\tan\dfrac{x}{2}=t とおくと dx=21+t2dtdx=\dfrac{2}{1+t^2}dt cos⁡x=1−t21+t2\cos x=\dfrac{1-t^2}{1+t^2}

x:0→π/2x:0\to\pi/2 t:0→1t:0\to1 これから

=∫0112−1−t21+t2⋅21+t2 dt=∫0122+2t2−1+t2 dt=∫0123t2+1 dt=∫0123t2+13 dt=23[11/3tan⁡−1t1/3]01=\int_0^1\frac{1}{2-\dfrac{1-t^2}{1+t^2}}\cdot\frac{2}{1+t^2}\,dt=\int_0^1\frac{2}{2+2t^2-1+t^2}\,dt=\int_0^1\frac{2}{3t^2+1}\,dt=\int_0^1\frac{\dfrac{2}{3}}{t^2+\dfrac{1}{3}}\,dt=\frac{2}{3}\left[\frac{1}{1/\sqrt3}\tan^{-1}\frac{t}{1/\sqrt3}\right]_0^1

=233[tan⁡−13 x]01=239π=\frac{2\sqrt3}{3}\left[\tan^{-1}\sqrt3\,x\right]_0^1=\frac{2\sqrt3}{9}\pi

(3)

∫0π/2dx4+5cos⁡x\displaystyle\int_0^{\pi/2}\frac{dx}{4+5\cos x} tan⁡x2=t\tan\dfrac{x}{2}=t とおくと dx=21+t2dtdx=\dfrac{2}{1+t^2}dt cos⁡x=1−t21+t2\cos x=\dfrac{1-t^2}{1+t^2}

x:0→π/2x:0\to\pi/2 t:0→1t:0\to1 これから

=∫0114+5⋅1−t21+t2⋅21+t2 dt=∫0124+4t2+5−5t2 dt=∫0129−t2 dt=−13[log⁡t−3t+3]01=−13log⁡12=\int_0^1\frac{1}{4+5\cdot\dfrac{1-t^2}{1+t^2}}\cdot\frac{2}{1+t^2}\,dt=\int_0^1\frac{2}{4+4t^2+5-5t^2}\,dt=\int_0^1\frac{2}{9-t^2}\,dt=-\frac{1}{3}\left[\log\frac{t-3}{t+3}\right]_0^1=-\frac{1}{3}\log\frac{1}{2}

=13log⁡2=\frac{1}{3}\log2

(4)

∫0π/4dx2+tan⁡x\int_0^{\pi/4} \frac{dx}{2+\tan x}

tan⁡x=t\tan x = t とおくと x=tan⁡−1tx = \tan^{-1} t dx=1t2+1dtdx = \dfrac{1}{t^2+1}dt

x:0→π/4x : 0 \to \pi/4 t:0→1t : 0 \to 1 これから

=∫0112+t⋅1t2+1dt=∫0115(1t+2+−t+2t2+1)dt=∫0115(1t+2−12⋅2tt2+1+2t2+1)dt=\int_0^1 \frac{1}{2+t}\cdot\frac{1}{t^2+1}dt = \int_0^1 \frac{1}{5}\left(\frac{1}{t+2}+\frac{-t+2}{t^2+1}\right)dt = \int_0^1 \frac{1}{5}\left(\frac{1}{t+2}-\frac{1}{2}\cdot\frac{2t}{t^2+1}+\frac{2}{t^2+1}\right)dt

=15[log⁡(t+2)−12log⁡(t2+1)+2tan⁡−1t]01=15(log⁡3−log⁡2−12log⁡2+π2)=π10+110log⁡98=\frac{1}{5}\left[\log(t+2)-\frac{1}{2}\log(t^2+1)+2\tan^{-1}t\right]_0^1 = \frac{1}{5}\left(\log 3-\log 2-\frac{1}{2}\log 2+\frac{\pi}{2}\right)=\frac{\pi}{10}+\frac{1}{10}\log\frac{9}{8}

解説: r31bn1z