【微分積分】4-3-4 漸化式の積分｜問題集
1.\(n\geqq2\)のとき、次の漸化式を証明しなさい。
(1)\(\displaystyle I_n=\int\frac{1}{(x^2+c^2)^n}dx\)とおくとき、\(\displaystyle I_n=\frac{1}{c^2}\left\{\frac{1}{2n-2}・\frac{x}{(x^2+c^2)^{n-1}}+\frac{2n-3}{2n-2}I_{n-1}\right\}\)を証明しなさい。
\(\displaystyle \left\{\frac{1}{(x^2+c^2)^{n-1}}\right\}'=-2(n-1)\frac{x}{(x^2+c^2)^n}\)より、
\(\displaystyle I_{n-1}=\frac{x}{(x^2+c^2)^{n-1}}+2(n-1)\int\frac{x^2}{(x^2+c^2)^n}dx\)
\(\displaystyle I_{n-1}=\frac{x}{(x^2+c^2)^{n-1}}+2(n-1)I_{n-1}-2(n-1)c^2I_n\)
\(I_n\)について解くと
\(\displaystyle I_n=\frac{1}{c^2}\left\{\frac{1}{2n-2}・\frac{x}{(x^2+c^2)^{n-1}}+\frac{2n-3}{2n-2}I_{n-1}\right\}\)
\(\displaystyle I_{n-1}=\frac{x}{(x^2+c^2)^{n-1}}+2(n-1)\int\frac{x^2}{(x^2+c^2)^n}dx\)
\(\displaystyle I_{n-1}=\frac{x}{(x^2+c^2)^{n-1}}+2(n-1)I_{n-1}-2(n-1)c^2I_n\)
\(I_n\)について解くと
\(\displaystyle I_n=\frac{1}{c^2}\left\{\frac{1}{2n-2}・\frac{x}{(x^2+c^2)^{n-1}}+\frac{2n-3}{2n-2}I_{n-1}\right\}\)
(2)\(\displaystyle I_n=\int\sin^nxdx\)とおくとき、\(\displaystyle I_n=-\frac{1}{n}\sin^{n-1}x\cos x+\frac{n-1}{n}I_{n-2}\)を証明しなさい。
\(\sin^n x=\sin^{n-2}x(1-\cos^2x)\)より
\(\displaystyle I_n=I_{n-2}-\int\sin^{n-2}x\cos^2xdx\)
\(\displaystyle I_n=I_{n-2}-\frac{1}{n-1}\int(\sin^{n-1}x)'\cos xdx\)
\(\displaystyle I_n=I_{n-2}-\frac{1}{n-1}\sin^{n-1}x\cos x-\frac{1}{n-1}I_n\)
\(I_n\)について解くと
\(\displaystyle I_n=-\frac{1}{n}\sin^{n-1}x\cos x+\frac{n-1}{n}I_{n-2}\)
\(\displaystyle I_n=I_{n-2}-\int\sin^{n-2}x\cos^2xdx\)
\(\displaystyle I_n=I_{n-2}-\frac{1}{n-1}\int(\sin^{n-1}x)'\cos xdx\)
\(\displaystyle I_n=I_{n-2}-\frac{1}{n-1}\sin^{n-1}x\cos x-\frac{1}{n-1}I_n\)
\(I_n\)について解くと
\(\displaystyle I_n=-\frac{1}{n}\sin^{n-1}x\cos x+\frac{n-1}{n}I_{n-2}\)
(3)\(\displaystyle I_n=\int x^n\sin xdx\)とおくとき、\(I_n=x^{n-1}(n\sin x-x\cos x)-n(n-1)I_{n-2}\)を証明しなさい。
\(\displaystyle I_n=\int x^n(-\cos x)'dx\)
\(\displaystyle =-x^n\cos x+n\int x^{n-1}\cos xdx\)
\(\displaystyle =-x^n\cos x+nx^{n-1}\sin x-n(n-1)\int x^{n-2}\sin xdx\)
\(\displaystyle =x^{n-1}(n\sin x-x\cos x)-n(n-1)I_{n-2}\)
\(\displaystyle =-x^n\cos x+n\int x^{n-1}\cos xdx\)
\(\displaystyle =-x^n\cos x+nx^{n-1}\sin x-n(n-1)\int x^{n-2}\sin xdx\)
\(\displaystyle =x^{n-1}(n\sin x-x\cos x)-n(n-1)I_{n-2}\)
(4)\(\displaystyle \int x^ne^{-x}dx=-x^ne^{-x}+n\int x^{n-1}e^{-x}dx\)を証明しなさい。
\(\displaystyle \int x^ne^{-x}dx\)
\(\displaystyle =x^n・(-e^{-x})'-\int(x^n)'・(-e^{-x})dx\)
\(\displaystyle =-x^ne^{-x}+n\int x^{n-1}e^{-x}dx\)
\(\displaystyle =x^n・(-e^{-x})'-\int(x^n)'・(-e^{-x})dx\)
\(\displaystyle =-x^ne^{-x}+n\int x^{n-1}e^{-x}dx\)
(5)\(\displaystyle \int\cos^nxdx\)
\(\displaystyle =\frac{1}{n}\cos^{n-1}x\sin x+\frac{n-1}{n}\int\cos^{n-2}xdx\)
を証明しなさい。
\(\displaystyle =\frac{1}{n}\cos^{n-1}x\sin x+\frac{n-1}{n}\int\cos^{n-2}xdx\)
を証明しなさい。
\(\displaystyle I_n=\int\cos^nxdx\)とおくとき
\(\displaystyle I_n=\cos^{n-1}x\sin x+(n-1)\int\sin^2x\cos^{n-2}xdx\)
\(\displaystyle I_n=\cos^{n-1}x\sin x+(n-1)\int\cos^{n-2}xdx\)
\(\displaystyle \ \ \ \ \ -(n-1)\int\cos^nxdx\)
\(\displaystyle I_n=\cos^{n-1}x\sin x+(n-1)\int\cos^{n-2}xdx\)
\(\displaystyle \ \ \ \ \ -(n-1)I_n\)
\(I_n\)について解くと
\(\displaystyle I_n=\frac{1}{n}\cos^{n-1}x\sin x+\frac{n-1}{n}\int\cos^{n-2}xdx\)
\(\displaystyle I_n=\cos^{n-1}x\sin x+(n-1)\int\cos^{n-2}xdx\)
\(\displaystyle \ \ \ \ \ -(n-1)\int\cos^nxdx\)
\(\displaystyle I_n=\cos^{n-1}x\sin x+(n-1)\int\cos^{n-2}xdx\)
\(\displaystyle \ \ \ \ \ -(n-1)I_n\)
\(I_n\)について解くと
\(\displaystyle I_n=\frac{1}{n}\cos^{n-1}x\sin x+\frac{n-1}{n}\int\cos^{n-2}xdx\)
(6)\(\displaystyle \int\tan^nxdx=\frac{1}{n-1}\tan^{n-1}x-\int\tan^{n-2}xdx\)を証明しなさい。
\(\displaystyle \int\tan^nxdx\)
\(\displaystyle =\int\tan^{n-2}x\tan^2xdx\)
\(\displaystyle =\int\tan^{n-2}x(\sec^2x-1)dx\)
\(\displaystyle =\int\tan^{n-2}x\sec^2xdx-\int\tan^{n-2}xdx\)
\(\displaystyle =\frac{1}{n-1}\tan^{n-1}x-\int\tan^{n-2}xdx\)
よって、
\(\displaystyle \int\tan^nxdx=\frac{1}{n-1}\tan^{n-1}x-\int\tan^{n-2}xdx\)
\(\displaystyle =\int\tan^{n-2}x\tan^2xdx\)
\(\displaystyle =\int\tan^{n-2}x(\sec^2x-1)dx\)
\(\displaystyle =\int\tan^{n-2}x\sec^2xdx-\int\tan^{n-2}xdx\)
\(\displaystyle =\frac{1}{n-1}\tan^{n-1}x-\int\tan^{n-2}xdx\)
よって、
\(\displaystyle \int\tan^nxdx=\frac{1}{n-1}\tan^{n-1}x-\int\tan^{n-2}xdx\)
(7)\(\displaystyle \int x^n\cos xdx=x^{n-1}(n\cos x+x\sin x)\)
\(\displaystyle \ \ \ \ \ -n(n-1)\int x^{n-2}\cos xdx\)
を証明しなさい。
\(\displaystyle \ \ \ \ \ -n(n-1)\int x^{n-2}\cos xdx\)
を証明しなさい。
\(\displaystyle I_n=\int x^n\cos xdx\)とおく。
\(\displaystyle I_n=x^n\sin x-n\int x^{n-1}\sin xdx\)
\(\displaystyle =x^n\sin x+nx^{n-1}\cos x-n(n-1)\int x^{n-2}\cos xdx\)
\(\displaystyle =x^{n-1}(n\cos x+x\sin x)-n(n-1)\int x^{n-2}\cos xdx\)
よって、
\(\displaystyle \int x^n\cos xdx=x^{n-1}(n\cos x+x\sin x)\)
\(\displaystyle \ \ \ \ \ -n(n-1)\int x^{n-2}\cos xdx\)
\(\displaystyle I_n=x^n\sin x-n\int x^{n-1}\sin xdx\)
\(\displaystyle =x^n\sin x+nx^{n-1}\cos x-n(n-1)\int x^{n-2}\cos xdx\)
\(\displaystyle =x^{n-1}(n\cos x+x\sin x)-n(n-1)\int x^{n-2}\cos xdx\)
よって、
\(\displaystyle \int x^n\cos xdx=x^{n-1}(n\cos x+x\sin x)\)
\(\displaystyle \ \ \ \ \ -n(n-1)\int x^{n-2}\cos xdx\)
(8)\(\displaystyle \int(\log x)^ndx=x(\log x)^n-n\int(\log x)^{n-1}dx\)を証明しなさい。
\(\displaystyle \int(\log x)^ndx\)
\(\displaystyle =x(\log x)^n-\int x・n(\log x)^{n-1}\frac{1}{x}dx\)
\(\displaystyle =x(\log x)^n-n\int(\log x)^{n-1}dx\)
よって、
\(\displaystyle \int(\log x)^ndx=x(\log x)^n-n\int(\log x)^{n-1}dx\)
\(\displaystyle =x(\log x)^n-\int x・n(\log x)^{n-1}\frac{1}{x}dx\)
\(\displaystyle =x(\log x)^n-n\int(\log x)^{n-1}dx\)
よって、
\(\displaystyle \int(\log x)^ndx=x(\log x)^n-n\int(\log x)^{n-1}dx\)
次の学習に進もう！