[mathjax]
積分問題39番
思考
根号を外すための置換を考えます。
\(1+\tan^2 \theta=\displaystyle\frac{1}{\cos^2 \theta}\) の関係から逆に考えていくと
\(x=\displaystyle\frac{1}{\cos \theta}\) とおけばうまくいくかもしれない、と考えられます。
\(dx=\displaystyle\frac{\sin \theta}{\cos^2 \theta}d\theta\) となります。
計算1
上の計算を問題の積分に代入すると
\(\displaystyle\int\sqrt{x^2-1}dx=\displaystyle\int\displaystyle\frac{\tan^2 \theta}{\cos \theta}d\theta\)
\(=\displaystyle\int\displaystyle\frac{1}{\cos \theta}\biggl(\displaystyle\frac{1}{\cos^2 \theta}-1\biggr)d\theta\)
\(=\displaystyle\int\displaystyle\frac{1}{\cos^3 \theta}d\theta\)\(-\displaystyle\int\displaystyle\frac{1}{\cos \theta}d\theta\)
第一項
\(=\displaystyle\int\displaystyle\frac{1}{\cos^3 \theta}d\theta\)
\(=\displaystyle\int\displaystyle\frac{\cos \theta}{\cos^4 \theta}d\theta\)
\(t=\sin \theta\) とおく。
\(\displaystyle\int\displaystyle\frac{dt}{(t^2-1)^2}\)
\(=\displaystyle\frac{1}{4}\displaystyle\int\biggl(\displaystyle\frac{1}{t+1}+\displaystyle\frac{1}{(t+1)^2}-\displaystyle\frac{1}{t-1}+\displaystyle\frac{1}{(t-1)^2}\biggr)dt\)
\(=\displaystyle\frac{1}{4}\log\biggl|\displaystyle\frac{t+1}{t-1}\biggr|-\displaystyle\frac{1}{4(t+1)}-\displaystyle\frac{1}{4(t-1)}+C\)
元に戻す。計算すると、
\(=\displaystyle\frac{1}{2}\log\biggl|\displaystyle\frac{1+\sin \theta}{\cos \theta}\biggr|+\displaystyle\frac{\sin \theta}{2\cos^2 \theta}+C\)
第二項
\(\displaystyle\int\displaystyle\frac{1}{\cos \theta}d\theta\)
\(t=\sin \theta\) とおく。
\(=\displaystyle\int\displaystyle\frac{dt}{1-t^2}\)
\(=\displaystyle\frac{1}{2}\log\biggl|\displaystyle\frac{1+t}{1-t}\biggr|+C\)
\(=\log\biggl|\displaystyle\frac{1+\sin \theta}{\cos \theta}\biggr|+C\)
合わせる
\(=\displaystyle\int\displaystyle\frac{1}{\cos^3 \theta}d\theta\)\(-\displaystyle\int\displaystyle\frac{1}{\cos \theta}d\theta\)
\(=-\displaystyle\frac{1}{2}\log\biggl|\displaystyle\frac{1+\sin \theta}{\cos \theta}\biggr|+\displaystyle\frac{\sin \theta}{2\cos^2 \theta}+C\)
\(=\displaystyle\frac{x}{2}\sqrt{x^2-1}\)\(-\displaystyle\frac{1}{2}\log|x+\sqrt{x^2-1}|+C\)
計算2
部分積分で解く。
\(\displaystyle\int\sqrt{x^2-1}dx\)
\(=x\sqrt{x^2-1}-\displaystyle\int\displaystyle\frac{x^2}{\sqrt{x^2-1}}dx\)
\(=x\sqrt{x^2-1}-\displaystyle\int\displaystyle\frac{(\sqrt{x^2-1})^2+1}{\sqrt{x^2-1}}dx\)
\(=x\sqrt{x^2-1}-\displaystyle\int\sqrt{x^2-1}dx-\displaystyle\int\displaystyle\frac{dx}{\sqrt{x^2-1}}\)
よって
\(\displaystyle\int\sqrt{x^2-1}dx=\displaystyle\frac{1}{2}x\sqrt{x^2-1}-\displaystyle\frac{1}{2}\log|x+\sqrt{x^2+1}|+C\)
答え
\(\displaystyle\frac{x}{2}\sqrt{x^2-1}\)\(-\displaystyle\frac{1}{2}\log|x+\sqrt{x^2-1}|+C\)
