【微分積分】4-2-3 部分積分法・置換積分法｜問題集
1.次の不定積分・定積分を求めなさい。
(1)\(\displaystyle \int\frac{2x+1}{x^2+x}dx\)
\(t=x^2+x\)とおくと、\(\displaystyle dx=\frac{1}{2x+1}dt\)
\(\displaystyle =\int\frac{1}{t}dt\)
\(\displaystyle =\log|t|+C\)
\(\displaystyle =\log|x^2+x|+C\)
\(\displaystyle =\int\frac{1}{t}dt\)
\(\displaystyle =\log|t|+C\)
\(\displaystyle =\log|x^2+x|+C\)
(2)\(\displaystyle \int x^2\sqrt{x+1}dx\)
\(t=x+1\)とおくと、\(\displaystyle dx=dt\)
\(\displaystyle =\int(t-1)^2\sqrt{t}dt\)
\(\displaystyle =\int(t^{\frac{5}{2}}-2t^{\frac{3}{2}}+t^{\frac{1}{2}})dt\)
\(\displaystyle =\frac{2}{7}t^{\frac{7}{2}}-\frac{4}{5}t^{\frac{5}{2}}+\frac{2}{3}t^{\frac{3}{2}}+C\)
\(\displaystyle =\frac{2}{7}(x+1)^{\frac{7}{2}}-\frac{4}{5}(x+1)^{\frac{5}{2}}+\frac{2}{3}(x+1)^{\frac{3}{2}}+C\)
\(\displaystyle =\int(t-1)^2\sqrt{t}dt\)
\(\displaystyle =\int(t^{\frac{5}{2}}-2t^{\frac{3}{2}}+t^{\frac{1}{2}})dt\)
\(\displaystyle =\frac{2}{7}t^{\frac{7}{2}}-\frac{4}{5}t^{\frac{5}{2}}+\frac{2}{3}t^{\frac{3}{2}}+C\)
\(\displaystyle =\frac{2}{7}(x+1)^{\frac{7}{2}}-\frac{4}{5}(x+1)^{\frac{5}{2}}+\frac{2}{3}(x+1)^{\frac{3}{2}}+C\)
(3)\(\displaystyle \int\frac{x}{\sqrt{x^2-4}}dx\)
\(t=x^2-4\)とおくと、\(\displaystyle dx=\frac{1}{2x}dt\)
\(\displaystyle =\int\frac{1}{2\sqrt{t}}dt\)
\(\displaystyle =\frac{1}{2}\int t^{-\frac{1}{2}}dt\)
\(\displaystyle =\frac{1}{2}・2t^{\frac{1}{2}}+C\)
\(\displaystyle =\sqrt{x^2-4}+C\)
\(\displaystyle =\int\frac{1}{2\sqrt{t}}dt\)
\(\displaystyle =\frac{1}{2}\int t^{-\frac{1}{2}}dt\)
\(\displaystyle =\frac{1}{2}・2t^{\frac{1}{2}}+C\)
\(\displaystyle =\sqrt{x^2-4}+C\)
(4)\(\displaystyle \int\sin2xdx\)
\(t=2x\)とおくと、\(\displaystyle dx=\frac{1}{2}dt\)
\(\displaystyle =\int\sin t・\frac{1}{2}dt\)
\(\displaystyle =\frac{1}{2}\int\sin tdt\)
\(\displaystyle =\frac{1}{2}・(-\cos t)+C\)
\(\displaystyle =-\frac{1}{2}\cos2x+C\)
\(\displaystyle =\int\sin t・\frac{1}{2}dt\)
\(\displaystyle =\frac{1}{2}\int\sin tdt\)
\(\displaystyle =\frac{1}{2}・(-\cos t)+C\)
\(\displaystyle =-\frac{1}{2}\cos2x+C\)
(5)\(\displaystyle \int\frac{x}{x^2+1}dx\)
\(t=x^2+1\)とおくと、\(\displaystyle dx=\frac{1}{2x}dt\)
\(\displaystyle =\int\frac{1}{2t}dt\)
\(\displaystyle =\frac{1}{2}\int\frac{1}{t}dt\)
\(\displaystyle =\frac{1}{2}\log|t|+C\)
\(\displaystyle =\frac{1}{2}\log(x^2+1)+C\)
\(\displaystyle =\int\frac{1}{2t}dt\)
\(\displaystyle =\frac{1}{2}\int\frac{1}{t}dt\)
\(\displaystyle =\frac{1}{2}\log|t|+C\)
\(\displaystyle =\frac{1}{2}\log(x^2+1)+C\)
(6)\(\displaystyle \int e^{2x}dx\)
\(t=2x\)とおくと、\(\displaystyle dx=\frac{1}{2}dt\)
\(\displaystyle =\int e^t\frac{1}{2}dt\)
\(\displaystyle =\frac{1}{2}\int e^tdt\)
\(\displaystyle =\frac{1}{2}e^t+C\)
\(\displaystyle =\frac{1}{2}e^{2x}+C\)
\(\displaystyle =\int e^t\frac{1}{2}dt\)
\(\displaystyle =\frac{1}{2}\int e^tdt\)
\(\displaystyle =\frac{1}{2}e^t+C\)
\(\displaystyle =\frac{1}{2}e^{2x}+C\)
(7)\(\displaystyle \int\frac{1}{x\log x}dx\)
\(t=\log x\)とおくと、\(\displaystyle dx=xdt\)
\(\displaystyle =\int \frac{1}{t}dt\)
\(\displaystyle =\log|t|+C\)
\(\displaystyle =\log|\log x|+C\)
\(\displaystyle =\int \frac{1}{t}dt\)
\(\displaystyle =\log|t|+C\)
\(\displaystyle =\log|\log x|+C\)
(8)\(\displaystyle \int xe^{x^2}dx\)
\(t=x^2\)とおくと、\(\displaystyle dx=\frac{1}{2x}dt\)
\(\displaystyle =\int e^t・\frac{1}{2}dt\)
\(\displaystyle =\frac{1}{2}\int e^tdt\)
\(\displaystyle =\frac{1}{2}e^t+C\)
\(\displaystyle =\frac{1}{2}e^{x^2}+C\)
\(\displaystyle =\int e^t・\frac{1}{2}dt\)
\(\displaystyle =\frac{1}{2}\int e^tdt\)
\(\displaystyle =\frac{1}{2}e^t+C\)
\(\displaystyle =\frac{1}{2}e^{x^2}+C\)
(9)\(\displaystyle \int \sin^2x\cos xdx\)
\(t=\sin x\)とおくと、\(\displaystyle dx=\frac{1}{\cos x}dt\)
\(\displaystyle =\int t^2dt\)
\(\displaystyle =\frac{1}{3}t^3+C\)
\(\displaystyle =\frac{\sin^3x}{3}+C\)
\(\displaystyle =\int t^2dt\)
\(\displaystyle =\frac{1}{3}t^3+C\)
\(\displaystyle =\frac{\sin^3x}{3}+C\)
(10)\(\displaystyle \int x\sqrt{1+x}dx\)
\(t=1+x\)とおくと、\(\displaystyle dx=dt\)
\(\displaystyle =\int(t-1)\sqrt{t}dt\)
\(\displaystyle =\int(t^{\frac{3}{2}}-t^{\frac{1}{2}})dt\)
\(\displaystyle =\frac{2}{5}t^{\frac{5}{2}}-\frac{2}{3}t^{\frac{3}{2}}+C\)
\(\displaystyle =\frac{2}{5}(1+x)^{\frac{5}{2}}-\frac{2}{3}(1+x)^{\frac{3}{2}}+C\)
\(\displaystyle =\int(t-1)\sqrt{t}dt\)
\(\displaystyle =\int(t^{\frac{3}{2}}-t^{\frac{1}{2}})dt\)
\(\displaystyle =\frac{2}{5}t^{\frac{5}{2}}-\frac{2}{3}t^{\frac{3}{2}}+C\)
\(\displaystyle =\frac{2}{5}(1+x)^{\frac{5}{2}}-\frac{2}{3}(1+x)^{\frac{3}{2}}+C\)
(11)\(\displaystyle \int\frac{\cos x-\sin x}{\sin x+\cos x}dx\)
\(t=\sin x+\cos x\)とおくと、\(\displaystyle dx=\frac{1}{\cos x-\sin x}dt\)
\(\displaystyle =\int\frac{1}{t}dt\)
\(\displaystyle =\log|t|+C\)
\(\displaystyle =\log|\sin x+\cos x|+C\)
\(\displaystyle =\int\frac{1}{t}dt\)
\(\displaystyle =\log|t|+C\)
\(\displaystyle =\log|\sin x+\cos x|+C\)
(12)\(\displaystyle \int\frac{e^x}{1+e^x}dx\)
\(t=1+e^x\)とおくと、\(\displaystyle dx=\frac{1}{e^x}dt\)
\(\displaystyle =\int\frac{1}{t}dt\)
\(\displaystyle =\log|t|+C\)
\(\displaystyle =\log(1+e^x)+C\)
\(\displaystyle =\int\frac{1}{t}dt\)
\(\displaystyle =\log|t|+C\)
\(\displaystyle =\log(1+e^x)+C\)
(13)\(\displaystyle \int\log xdx\)
\(\displaystyle =x\log x-\int x・\frac{1}{x}dx\)
\(\displaystyle =x\log x-x+C\)
\(\displaystyle =x(\log x-1)+C\)
\(\displaystyle =x\log x-x+C\)
\(\displaystyle =x(\log x-1)+C\)
(14)\(\displaystyle \int xe^{-x}dx\)
\(\displaystyle =x・(-e^{-x})-\int-e^{-x}dx\)
\(\displaystyle =-xe^{-x}-e^{-x}+C\)
\(\displaystyle =-e^{-x}(x+1)+C\)
\(\displaystyle =-xe^{-x}-e^{-x}+C\)
\(\displaystyle =-e^{-x}(x+1)+C\)
(15)\(\displaystyle \int x\sin xdx\)
\(\displaystyle =x・(-\cos x)-\int-\cos xdx\)
\(\displaystyle =-x\cos x+\sin x+C\)
\(\displaystyle =-x\cos x+\sin x+C\)
(16)\(\displaystyle \int xe^{2x}dx\)
\(\displaystyle =x・\frac{1}{2}e^{2x}-\int\frac{1}{2}e^{2x}dx\)
\(\displaystyle =\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+C\)
\(\displaystyle =\frac{1}{2}e^{2x}\left(x-\frac{1}{2}\right)+C\)
\(\displaystyle =\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}+C\)
\(\displaystyle =\frac{1}{2}e^{2x}\left(x-\frac{1}{2}\right)+C\)
(17)\(\displaystyle \int x^2e^xdx\)
\(\displaystyle =x^2e^x-\int 2xe^xdx\)
\(\displaystyle =x^2e^x-2\left(xe^x-\int e^xdx\right)+C\)
\(\displaystyle =x^2e^x-2xe^x+2e^x+C\)
\(\displaystyle =e^x(x^2-2x+2)+C\)
\(\displaystyle =x^2e^x-2\left(xe^x-\int e^xdx\right)+C\)
\(\displaystyle =x^2e^x-2xe^x+2e^x+C\)
\(\displaystyle =e^x(x^2-2x+2)+C\)
(18)\(\displaystyle \int x^2\sin xdx\)
\(\displaystyle =-x^2\cos x-\int 2x・(-\cos x)dx\)
\(\displaystyle =-x^2\cos x+2\left(x\sin x-\int\sin xdx\right)+C\)
\(\displaystyle =-x^2\cos x+2(x\sin x+\cos x)+C\)
\(\displaystyle =2x\sin x+(2-x^2)\cos x+C\)
\(\displaystyle =-x^2\cos x+2\left(x\sin x-\int\sin xdx\right)+C\)
\(\displaystyle =-x^2\cos x+2(x\sin x+\cos x)+C\)
\(\displaystyle =2x\sin x+(2-x^2)\cos x+C\)
(19)\(\displaystyle \int x^5e^{x^3}dx\)
\(\displaystyle =\int x^3・x^2e^{x^3}dx\)
\(\displaystyle =\frac{1}{3}x^3e^{x^3}-\int x^2e^{x^3}dx\)
\(\displaystyle =\frac{1}{3}x^3e^{x^3}-\frac{1}{3}e^{x^3}+C\)
\(\displaystyle =\frac{1}{3}e^{x^3}(x^3-1)+C\)
\(\displaystyle =\frac{1}{3}x^3e^{x^3}-\int x^2e^{x^3}dx\)
\(\displaystyle =\frac{1}{3}x^3e^{x^3}-\frac{1}{3}e^{x^3}+C\)
\(\displaystyle =\frac{1}{3}e^{x^3}(x^3-1)+C\)
(20)\(\displaystyle \int x\cos xdx\)
\(\displaystyle =x\sin x-\int\sin xdx\)
\(\displaystyle =x\sin x+\cos x+C\)
\(\displaystyle =x\sin x+\cos x+C\)
(21)\(\displaystyle \int_0^13x^2(x^3+1)^4dx\)
\(t=x^3+1\)とおくと、\(\displaystyle dx=\frac{1}{3x^2}dt\)
\(\displaystyle =\int_1^2t^4dt\)
\(\displaystyle =\left[\frac{t^5}{5}\right]_1^2\)
\(\displaystyle =\frac{32-1}{5}\)
\(\displaystyle =\frac{31}{5}\)
\(\displaystyle =\int_1^2t^4dt\)
\(\displaystyle =\left[\frac{t^5}{5}\right]_1^2\)
\(\displaystyle =\frac{32-1}{5}\)
\(\displaystyle =\frac{31}{5}\)
(22)\(\displaystyle \int_{-2}^2\sin xdx\)
\(\sin x\)は奇関数なので、
\(=0\)
\(=0\)
(23)\(\displaystyle \int_{-1}^1\sin^3xdx\)
\(\sin^3x\)は奇関数なので、
\(=0\)
\(=0\)
次の学習に進もう！