[mathjax]
積分問題37番
思考
\(x^2+1\)があるので\(\tan \theta\)で置き換えてみます。
計算1
\(x=\tan \theta\) と置きます。
\(dx=\displaystyle\frac{d\theta}{\cos^2 \theta}\)
\(=\displaystyle\int_{0}^{\frac{\pi}{4}} \tan^3 \theta \cos \theta d\theta\)
\(=\displaystyle\int_{0}^{\frac{\pi}{4}}\displaystyle\frac{\sin^3 \theta}{\cos^2 \theta}d\theta\)
\(t=\cos \theta\) とおくと、\(dt=-\sin \theta d\theta\) なので
\(=\displaystyle\int_{1}^{\frac{1}{\sqrt2}}\biggl(1-\displaystyle\frac{1}{t^2}\biggr)dt\)
\(=\biggl[t+\displaystyle\frac{1}{t}\biggr]_{1}^{\frac{1}{\sqrt2}}\)
\(=\displaystyle\frac{3\sqrt2}{2}-2\)
計算2
分子を、\(x^3=x(x^2+1)-x\) と変形する。
\(\displaystyle\int_{0}^{1} \displaystyle\frac{x^3}{(x^2+1)^{\frac{3}{2}}}dx\)
\(=\displaystyle\int_{0}^{1} \displaystyle\frac{x(x^2+1)-x}{(x^2+1)^{\frac{3}{2}}}dx\)
\(=\displaystyle\int_{0}^{1} \displaystyle\frac{x}{\sqrt{x^2+1}}-\displaystyle\int_{0}^{1} \displaystyle\frac{x}{(x^2+1)^{\frac{3}{2}}}dx\)
\(=\biggl[\sqrt{x^2+1}\biggr]_{0}^{1}+\biggl[\displaystyle\frac{1}{\sqrt{x^2+1}}\biggr]_{0}^{1}\)
\(=\displaystyle\frac{3\sqrt2}{2}-2\)
答え
\(\displaystyle\frac{3\sqrt2}{2}-2\)
