【微分積分】4-5-3 広義の置換積分｜要点まとめ

\(\displaystyle =\lim_{t\to\infty}\int_0^t\frac{1}{e^x+1}dx\)
\(u=e^x\)とおくと、\(\displaystyle dx=\frac{1}{u}du\)
\(\displaystyle =\lim_{t\to\infty}\int_1^t\left(\frac{1}{u}-\frac{1}{u+1}\right)du\)
\(\displaystyle =\lim_{t\to\infty}\left[\log\left|\frac{u}{u+1}\right|\right]_1^t\)
\(\displaystyle =\lim_{t\to\infty}\log\frac{2t}{t+1}\)
\(\displaystyle =\log2\)

\(\displaystyle =\int_{-\infty}^\infty\frac{2e^x}{e^{2x}+1}dx\)
\(\displaystyle =\int_{-\infty}^0\frac{2e^x}{e^{2x}+1}dx+\int_0^\infty\frac{2e^x}{e^{2x}+1}dx\)
\(\displaystyle =\lim_{t\to-\infty}\int_t^0\frac{2e^x}{e^{2x}+1}dx+\lim_{t\to\infty}\int_0^t\frac{2e^x}{e^{2x}+1}dx\)
\(u=e^x\)とおくと、\(\displaystyle dx=\frac{1}{e^x}du\)
\(\displaystyle =\lim_{t\to-\infty}\int_t^0\frac{2}{u^2+1}du+\lim_{t\to\infty}\int_0^t\frac{2}{u^2+1}dx\)
\(\displaystyle =\lim_{t\to-\infty}[2\tan^{-1}e^x]_t^0+\lim_{t\to\infty}[2\tan^{-1}e^x]_0^t\)
\(\displaystyle =0+2・\frac{\pi}{2}\)
\(\displaystyle =\pi\)

\(u=\sin^{-1}x\)とおくと、\(\displaystyle dx=\sqrt{1-x^2}du\)
\(\displaystyle =\int_0^{\frac{\pi}{2}}u\sin udu\)
\(\displaystyle =[-u\cos u+\sin u]_0^{\frac{\pi}{2}}\)
\(\displaystyle =1\)