積分問題bot112

シェアする

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

\(\displaystyle\frac{3\sqrt{2}}{2}-1-\displaystyle\frac{\pi}{4}\)

シェアする