置換積分法
【置換積分法】
\(x=g(t)\)とおくとき、
\(\displaystyle \int f(x)dx=\int f(g(t))g'(t)dt\)
【例題】次の不定積分を求めなさい。
(1)\(\displaystyle \int\sqrt[3]{2x+1}dx\)
\(2x+1=t\)とおくと、\(\displaystyle dx=\frac{1}{2}dt\)なので、
\(\displaystyle =\int\sqrt[3]{t}・\frac{1}{2}dt\)
\(\displaystyle =\frac{1}{2}\int t^{\frac{1}{3}}dt\)
\(\displaystyle =\frac{3}{8}t^{\frac{4}{3}}+C\)
\(\displaystyle =\frac{3}{8}(2x+1)\sqrt[3]{2x+1}+C\ \ \)(\(C\)は積分定数)
(2)\(\displaystyle \int\sin(3x-2)dx\)
\(3x-2=t\)とおくと、\(\displaystyle dx=\frac{1}{3}dt\)なので、
\(\displaystyle =\int\sin t・\frac{1}{3}dt\)
\(\displaystyle =\frac{1}{3}\int\sin tdt\)
\(\displaystyle =-\frac{1}{3}\cos t+C\)
\(\displaystyle =-\frac{1}{3}\cos(3x-2)+C\ \ \)(\(C\)は積分定数)
(3)\(\displaystyle \int\frac{x}{(x+1)^2}dx\)
\(x+1=t\)とおくと、\(\displaystyle dx=dt\)なので、
\(\displaystyle =\int\frac{t-1}{t^2}dt\)
\(\displaystyle =\int\left(\frac{1}{t}-\frac{1}{t^2}\right)dt\)
\(\displaystyle =\log|t|+\frac{1}{t}+C\)
\(\displaystyle =\log|x+1|+\frac{1}{x+1}+C\ \ \)(\(C\)は積分定数)
(4)\(\displaystyle \int x\sqrt{x-3}dx\)
\(x-3=t\)とおくと、\(\displaystyle dx=dt\)なので、
\(\displaystyle =\int(t+3)\sqrt{t}dt\)
\(\displaystyle =\int(t^{\frac{3}{2}}+3t^{\frac{1}{2}})dt\)
\(\displaystyle =\frac{2}{5}t^{\frac{5}{2}}+2t^{\frac{3}{2}}+C\)
\(\displaystyle =\frac{2}{5}t^{\frac{3}{2}}(t+5)+C\)
\(\displaystyle =\frac{2}{5}(x-3)(x+2)\sqrt{x-3}+C\ \ \)(\(C\)は積分定数)
(5)\(\displaystyle \int2xe^{x^2}dx\)
\(x^2=t\)とおくと、\(\displaystyle 2xdx=dt\)なので、
\(\displaystyle =\int e^tdt\)
\(\displaystyle =e^t+C\)
\(\displaystyle =e^{x^2}+C\ \ \)(\(C\)は積分定数)
(6)\(\displaystyle \int\sin^3x\cos xdx\)
\(\sin x=t\)とおくと、\(\cos xdx=dt\)なので、
\(\displaystyle =\int t^3dt\)
\(\displaystyle =\frac{1}{4}t^4+C\)
\(\displaystyle =\frac{1}{4}\sin^4x+C\ \ \)(\(C\)は積分定数)
(7)\(\displaystyle \int\frac{2x}{\sqrt{x^2+2}}dx\)
\(x^2+2=t\)とおくと、\(2xdx=dt\)なので、
\(\displaystyle =\int t^{-\frac{1}{2}}dt\)
\(\displaystyle =2\sqrt{t}+C\)
\(\displaystyle =2\sqrt{x^2+2}+C\ \ \)(\(C\)は積分定数)
(8)\(\displaystyle \int\frac{2x}{x^2+2}dx\)
\(x^2+2=t\)とおくと、\(2xdx=dt\)なので、
\(\displaystyle =\int t^{-1}dt\)
\(\displaystyle =\log{|t|}+C\)
\(\displaystyle =\log{(x^2+2)}+C\ \ \)(\(C\)は積分定数)
(9)\(\displaystyle \int\tan xdx\)
\(\cos x=t\)とおくと、\(-\sin xdx=dt\)なので、
\(\displaystyle =\int \frac{-1}{t}dt\)
\(\displaystyle =-\log{|t|}+C\)
\(\displaystyle =-\log{|\cos x|}+C\ \ \)(\(C\)は積分定数)
部分積分法
【部分積分法】
\(\displaystyle \int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx\)
【例題】次の不定積分を求めなさい。
(1)\(\displaystyle \int xe^xdx\)
\(\displaystyle =xe^x-\int(x)'e^xdx\)
\(\displaystyle =xe^x-\int e^xdx\)
\(\displaystyle =xe^x-e^x+C\ \ \)(\(C\)は積分定数)
(2)\(\displaystyle \int\log{x}dx\)
\(\displaystyle =x\log{x}-\int(\log{x})'xdx\)
\(\displaystyle =x\log{x}-\int dx\)
\(\displaystyle =x\log{x}-x+C\ \ \)(\(C\)は積分定数)
(3)\(\displaystyle \int x^2\sin xdx\)
\(\displaystyle =-x^2\cos x-\int(x^2)'(-\cos x)dx\)
\(\displaystyle =-x^2\cos x+2\int x\cos xdx\)
\(\displaystyle =-x^2\cos x+2\int x(\sin x)'dx\)
\(\displaystyle =-x^2\cos x+2\{x\sin x-\int (x)'\sin xdx\}\)
\(\displaystyle =-x^2\cos x+2x\sin x+2\cos x+C\ \ \)(\(C\)は積分定数)