【高校数学Ⅲ】5-2-2 置換積分法と部分積分法｜問題集

\(1-x=t\)とおくと、\(dx=-dt\)なので、
\(\displaystyle =\int_1^0 (1-t)t^5・-dt\)
\(\displaystyle =\int_0^1(t^5+t^6)dt\)
\(\displaystyle =\left[\frac{1}{6}t^6-\frac{1}{7}t^7\right]_0^1\)
\(\displaystyle =\frac{1}{6}-\frac{1}{7}\)
\(\displaystyle =\frac{1}{42}\)

\(x+3=t\)とおくと、\(dx=dt\)なので、
\(\displaystyle =\int_0^1 (t-3)t^4dt\)
\(\displaystyle =\int_0^1(t^5-3t^4)dt\)
\(\displaystyle =\left[\frac{1}{6}t^6-\frac{3}{5}t^5\right]_0^1\)
\(\displaystyle =\frac{1}{6}-\frac{3}{5}\)
\(\displaystyle =-\frac{13}{30}\)

\(\sqrt{x-1}=t\)とおくと、\(dx=2tdt\)なので、
\(\displaystyle =2\int_1^2(t^2+1)t^2dt\)
\(\displaystyle =2\int_1^2(t^4+t^2)dt\)
\(\displaystyle =2\left[\frac{1}{5}t^5+\frac{1}{3}t^3\right]_1^2\)
\(\displaystyle =2\left(\frac{32}{5}+\frac{8}{3}-\frac{1}{5}-\frac{1}{3}\right)\)
\(\displaystyle =\frac{256}{15}\)

\(x+2=t\)とおくと、\(dx=dt\)なので、
\(\displaystyle =\int_1^4 \frac{t-2}{\sqrt{t}}dt\)
\(\displaystyle =\int_1^4(t^{\frac{1}{2}}-2t^{-\frac{1}{2}})dt\)
\(\displaystyle =\left[\frac{2}{3}t^{\frac{3}{2}}-4t^{\frac{1}{2}}\right]_1^4\)
\(\displaystyle =\left(\frac{2}{3}・4^{\frac{3}{2}}-4・4^{\frac{1}{2}}-\frac{2}{3}+4\right)\)
\(\displaystyle =\frac{2}{3}\)

\(x=\sin\theta\)とおくと、\(dx=\cos\theta d\theta\)なので、
\(\displaystyle =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1-\sin^2\theta}\cos\theta d\theta\)
\(\displaystyle =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2\theta d\theta\)
\(\displaystyle =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos2\theta}{2}d\theta\)
\(\displaystyle =\frac{1}{2}\left[\theta+\frac{1}{2}\sin2\theta\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\)
\(\displaystyle =\frac{1}{2}\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\)
\(\displaystyle =\frac{\pi}{2}\)

\(x=2\sin\theta\)とおくと、\(dx=2\cos\theta d\theta\)なので、
\(\displaystyle =2\int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{4-4\sin^2\theta}\cos\theta d\theta\)
\(\displaystyle =4\int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2\theta d\theta\)
\(\displaystyle =2\int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} (1+\cos2\theta)d\theta\)
\(\displaystyle =2\left[\theta+\frac{1}{2}\sin2\theta\right]_{-\frac{\pi}{6}}^{\frac{\pi}{3}}\)
\(\displaystyle =2\left(\frac{\pi}{3}+\frac{\sqrt{3}}{4}+\frac{\pi}{6}+\frac{\sqrt{3}}{4}\right)\)
\(\displaystyle =\pi+\sqrt{3}\)

\(x=2\sin\theta\)とおくと、\(dx=2\cos\theta d\theta\)なので、
\(\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{2\cos\theta}{\sqrt{4-4\sin^2\theta}}d\theta\)
\(\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d\theta\)
\(\displaystyle =[\theta]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\)
\(\displaystyle =\frac{\pi}{3}-\frac{\pi}{6}\)
\(\displaystyle =\frac{\pi}{6}\)

\(x=3\tan\theta\)とおくと、\(\displaystyle dx=\frac{3}{\cos^2\theta}d\theta\)なので、
\(\displaystyle =\int_0^{\frac{\pi}{3}} \frac{\cos^2\theta}{9}・\frac{3}{\cos^2\theta}d\theta\)
\(\displaystyle =\frac{1}{3}\int_0^{\frac{\pi}{3}} d\theta\)
\(\displaystyle =[\theta]_0^{\frac{\pi}{3}}\)
\(\displaystyle =\frac{\pi}{9}\)

\(x=2\tan\theta\)とおくと、\(\displaystyle dx=\frac{2}{\cos^2\theta}d\theta\)なので、
\(\displaystyle =\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos^2\theta}{4}・\frac{2}{\cos^2\theta}d\theta\)
\(\displaystyle =\frac{1}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} d\theta\)
\(\displaystyle =\frac{1}{2}[\theta]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\)
\(\displaystyle =\frac{\pi}{4}\)

\(\displaystyle =2\int_0^2 (3x^2+5)dx\)
\(\displaystyle =2[x^3+5x]_0^2\)
\(\displaystyle =2・18\) \(\displaystyle =36\)

\(=0\)

\(=0\)

\(\displaystyle =\frac{1}{2}・2\int_0^{\frac{\pi}{2}} (1-\cos2x)dx\)
\(\displaystyle =\left[(x-\frac{1}{2}\sin2x)\right]_0^{\frac{\pi}{2}}\)
\(\displaystyle =\frac{\pi}{2}\)

\(=0\)

\(\displaystyle =\int_{-1}^1 (x^5-3x^4+3x^3-x^2)dx\)
\(\displaystyle =2\int_0^1 (-3x^4-x^2)dx\)
\(\displaystyle =2\left[-\frac{3}{5}x^5-\frac{1}{3}x^3\right]_0^2\)
\(\displaystyle =-\frac{6}{5}-\frac{2}{3}\)
\(\displaystyle =-\frac{28}{15}\)

\(\displaystyle =\int_1^e (x)'\log xdx\)
\(\displaystyle =[x\log x]_1^e-\int_1^e x(\log x)'dx\)
\(\displaystyle =e-0-\int_1^e dx\)
\(\displaystyle =e-[x]_1^e\)
\(=e-(e-1)\)
\(=1\)

\(\displaystyle =\int_{-\pi}^{\pi} x^2(\sin x)'dx\)
\(\displaystyle =[x^2\sin x]_{-\pi}^{\pi}-\int_{-\pi}^{\pi} (x^2)'\sin xdx\)
\(\displaystyle =0-2\int_{-\pi}^{\pi} x\sin xdx\)
\(\displaystyle =-2\int_{-\pi}^{\pi} x(-\cos x)'dx\)
\(\displaystyle =[2x\cos x]_{-\pi}^{\pi}-\int_{-\pi}^{\pi} 2\cos xdx\)
\(\displaystyle =-4\pi-[2\sin x]_{-\pi}^{\pi}\)
\(=-4\pi\)

\(\displaystyle =\int_0^\pi x(-\cos x)'dx\)
\(\displaystyle =[-x\cos x]_0^\pi+\int_0^\pi \cos xdx\)
\(\displaystyle =\pi+[\sin x]_0^\pi\)
\(=\pi\)

\(\displaystyle =\int_0^1 x(e^x)'dx\)
\(\displaystyle =[xe^x]_0^1-\int_0^1 e^xdx\)
\(\displaystyle =e-[e^x]_0^1\)
\(=1\)

\(\displaystyle =\int_1^2 (\frac{1}{2}x^2)'\log xdx\)
\(\displaystyle =\frac{1}{2}[x^2\log x]_1^2-\frac{1}{2}\int_1^2 x^2・\frac{1}{x}dx\)
\(\displaystyle =\frac{1}{2}・4\log2-\frac{1}{4}[x^2]_1^2\)
\(\displaystyle =2\log2-\frac{3}{4}\)

\(\displaystyle =\int_{-1}^1 (\frac{1}{4}(x+1)^4)'(x-1)dx\)
\(\displaystyle =\frac{1}{4}[(x+1)^4(x-1)]_{-1}^1-\frac{1}{4}\int_{-1}^1 (x+1)^4dx\)
\(\displaystyle =0-0-\frac{1}{20}[(x+1)^5]_{-1}^1\)
\(\displaystyle =-\frac{1}{20}・2^5\)
\(\displaystyle =-\frac{8}{5}\)

\(\displaystyle =\int_0^{\frac{\pi}{2}} (e^x)'\cos xdx\)
\(\displaystyle =[e^x\cos x]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}} e^x\sin xdx\)
\(\displaystyle =-1+\int_0^{\frac{\pi}{2}} (e^x)'\sin xdx\)
\(\displaystyle =-1+[e^x\sin x]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} e^x\cos xdx\)
\(\displaystyle \int_0^{\frac{\pi}{2}} e^x\cos xdx=A\)とおくと、
\(A=-1+e^{\frac{\pi}{2}}-A\)
\(2A=-1+e^{\frac{\pi}{2}}\)
\(\displaystyle A=\frac{e^{\frac{\pi}{2}}-1}{2}\)

\(\displaystyle I_0=\int_0^{\frac{\pi}{2}}dx=\frac{\pi}{2}\)
\(\displaystyle I_1=\int_0^{\frac{\pi}{2}}\sin xdx=-[\cos x]_0^{\frac{\pi}{2}}=1\)

\(\displaystyle I_n=\int_0^{\frac{\pi}{2}} \sin^{n-1}x(-\cos x)'dx\)
\(\displaystyle =-[\sin^{n-1}x\cos x]_0^{\frac{\pi}{2}}+(n-1)\int_0^{\frac{\pi}{2}} \sin^{n-2}x\cos^2 xdx\)
\(\displaystyle =(n-1)\int_0^{\frac{\pi}{2}} \sin^{n-2}x(1-\sin^2x)dx\)
\(\displaystyle =(n-1)\int_0^{\frac{\pi}{2}} (\sin^{n-2}x-\sin^nx)dx\)
\(\displaystyle =(n-1)\{\int_0^{\frac{\pi}{2}} \sin^{n-2}xdx-\int_0^{\frac{\pi}{2}} \sin^nxdx\}\)
\(=(n-1)(I_{n-2}-I_n)\)