[mathjax]
積分問題13番
問題
解答
まずは分子を分割する。
\(\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \displaystyle\frac{\cos x+1}{\tan^2 x}dx\)
\(=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \biggl(\displaystyle\frac{\cos^3 x}{\sin^2 x}+\displaystyle\frac{1}{\tan^2 x}\biggr)dx\)
\(=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\displaystyle\frac{(1-\sin^2 x)\cos x}{\sin^2 x}dx\)\(+\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \biggl(\displaystyle\frac{1}{\sin^2 x}-1\biggr)dx\)
第一項
\(t=\sin x\)とおく。
\(\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\displaystyle\frac{(1-\sin^2 x)\cos x}{\sin^2 x}dx\)
\(=\displaystyle\int_{\frac{1}{\sqrt{2}}}^{1}\displaystyle\frac{(1-t^2)}{t^2}dt\)
\(=\displaystyle\int_{\frac{1}{\sqrt{2}}}^{1}\biggl(\displaystyle\frac{1}{t^2}-1\biggr)dt\)
\(=\biggl[-\displaystyle\frac{1}{t}-t\biggr]_{\frac{1}{\sqrt{2}}}^{1}\)
\(=\displaystyle\frac{3\sqrt{2}}{2}-2\)
第二項
\(\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \biggl(\displaystyle\frac{1}{\sin^2 x}-1\biggr)dx\)
\(=\biggl[-\displaystyle\frac{1}{\tan x}-x\biggr]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \)
\(=1-\displaystyle\frac{\pi}{4}\)
まとめ
第一項と第二項をまとめると以下のようになる
\(\displaystyle\frac{3\sqrt{2}}{2}-1-\displaystyle\frac{\pi}{4}\)
答え
\(\displaystyle\frac{3\sqrt{2}}{2}-1-\displaystyle\frac{\pi}{4}\)
