【2012京都大学】部分積分、tanθの置換積分｜数学Ⅲ：積分の計算

【2012京都大学】

定積分 \(\displaystyle\int^{\sqrt{3}}_{1}\displaystyle\frac{1}{x^2}\log{\sqrt{1+x^2}} dx\) の値を求めよ．

解答・解説

\(\displaystyle\int^{\sqrt{3}}_{1}\displaystyle\frac{1}{x^2}\log{\sqrt{1+x^2}} dx\)

\(=\displaystyle\frac{1}{2}\displaystyle\int^{\sqrt{3}}_{1}x^{-2}\log(1+x^2) dx\)

\(\displaystyle\int f(x)g^{\prime}(x)\enspace dx=f(x)g(x)-\displaystyle\int f^{\prime}(x)g(x)\enspace dx\)

\(=\displaystyle\frac{1}{2}\left\{\Bigl[-\displaystyle\frac{1}{x}\log(1+x^2)\Bigr]^{\sqrt{3}}_{1}-\displaystyle\int^{\sqrt{3}}_{1}\left(-\displaystyle\frac{1}{x}\right)\cdot\displaystyle\frac{2x}{1+x^2} dx\right\}\)

\(=-\displaystyle\frac{1}{2}\Bigl[\displaystyle\frac{1}{x}\log(1+x^2)\Bigr]^{\sqrt{3}}_{1}+\displaystyle\int^{\sqrt{3}}_{1}\displaystyle\frac{1}{1+x^2} dx\)

ここで，

\(-\displaystyle\frac{1}{2}\Bigl[\displaystyle\frac{1}{x}\log(1+x^2)\Bigr]^{\sqrt{3}}_{1}=-\displaystyle\frac{1}{2}\left(\displaystyle\frac{1}{\sqrt{3}}\log4-\log2\right)=\left(\displaystyle\frac{1}{2}-\displaystyle\frac{1}{\sqrt{3}}\right)\log2\)

また，\(\displaystyle\int^{\sqrt{3}}_{1}\displaystyle\frac{1}{1+x^2} dx\) について

\(\displaystyle\int \displaystyle\frac{1}{x^2+a^2}\enspace dx\)

☞　\(x=a\tan\theta\) と置換する

\(x=\tan \theta\) とおくと，

\(\displaystyle\frac{dx}{d \theta}=\displaystyle\frac{1}{\cos^2 \theta}\) ,

\(x\)：\(1 \rightarrow \sqrt{3}\) のとき \(\theta\)：\(\displaystyle\frac{\pi}{4} \rightarrow \displaystyle\frac{\pi}{3}\) より

\(\displaystyle\int^{\sqrt{3}}_{1}\displaystyle\frac{1}{1+x^2} dx\)

\(=\displaystyle\int^{\frac{\pi}{3}}_{\frac{\pi}{4}}\displaystyle\frac{1}{1+\tan^2\theta}\cdot\displaystyle\frac{1}{\cos^2\theta} d \theta\)

\(=\displaystyle\int^{\frac{\pi}{3}}_{\frac{\pi}{4}}d \theta=\Bigl[ \theta\Bigr]^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\displaystyle\frac{\pi}{12}\)

したがって，

(与式)\(=\left(\displaystyle\frac{1}{2}-\displaystyle\frac{1}{\sqrt{3}}\right)\log2+\displaystyle\frac{\pi}{12}\)

\(=\displaystyle\frac{\pi}{12}+\displaystyle\frac{3-2\sqrt{3}}{6}\log2\)