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積分問題8番
思考
有名な積分です。
まずは部分積分を行います。（\(x\)の微分がかけられていると考えて）
\(x=\tan\theta\)と置き換えても解ける。
計算1
\(\displaystyle\int \sqrt{1+x^2} dx\)
\(=x\sqrt{1+x^2}-\displaystyle\int\displaystyle\frac{x^2}{\sqrt{x^2+1}}dx\)
\(=x\sqrt{1+x^2}-\displaystyle\int\displaystyle\frac{x^2+1-1}{\sqrt{x^2+1}}dx\)
\(=x\sqrt{1+x^2}-\)\(\displaystyle\int\sqrt{1+x^2}dx\)\(+\displaystyle\int\displaystyle\frac{dx}{\sqrt{x^2+1}}\)
よって
\(2\displaystyle\int \sqrt{1+x^2} dx\)\(=x\sqrt{x^2+1}+\log(x+\sqrt{x^2+1})+C\)
※ 最後の\(\displaystyle\int\displaystyle\frac{dx}{\sqrt{x^2+1}}\)の積分。
\(t=x+\sqrt{x^2+1}\) とおくと（やや唐突ですが）
\(\displaystyle\frac{dt}{dx}=1+\displaystyle\frac{x}{\sqrt{x^2+1}}=\displaystyle\frac{t}{\sqrt{x^2+1}}\)
よって
\(\displaystyle\int\displaystyle\frac{dx}{\sqrt{x^2+1}}=\displaystyle\int\displaystyle\frac{dt}{t}\)
\(=\log t+C=\log(x+\sqrt{x^2+1})+C\)
計算2
\(\displaystyle\int \sqrt{1+x^2} dx\)
\(=\displaystyle\int\displaystyle\frac{d\theta}{\cos^3 \theta}\) \(x=\tan\theta\)
\(=\displaystyle\int\displaystyle\frac{\cos \theta}{\cos^4 \theta}d\theta\)
\(=\displaystyle\int\displaystyle\frac{dt}{(t^2-1)^2}\) \(t=\sin \theta\) とした
\(=\displaystyle\frac{1}{4}\displaystyle\int\biggl(\displaystyle\frac{1}{t+1}+\displaystyle\frac{1}{(t+1)^2}-\displaystyle\frac{1}{t-1}+\displaystyle\frac{1}{(t-1)^2}\biggr)dt\)
\(=\displaystyle\frac{1}{4}\log\biggl|\displaystyle\frac{t+1}{t-1}\biggr|-\displaystyle\frac{1}{4(t+1)}-\displaystyle\frac{1}{4(t-1)}+C\)
\(=\displaystyle\frac{1}{2}\log\biggl|\displaystyle\frac{1+\sin \theta}{\cos \theta}\biggr|+\displaystyle\frac{\sin \theta}{2\cos^2 \theta}+C\)
\(=\displaystyle\frac{1}{2}x\sqrt{x^2+1}+\displaystyle\frac{1}{2}\log(x+\sqrt{x^2+1})+C\)
答え
\(\displaystyle\frac{1}{2}x\sqrt{x^2+1}+\displaystyle\frac{1}{2}\log(x+\sqrt{x^2+1})+C\)
