積分問題bot104

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問題

 

計算

本題の計算に入る前に、一般的な話をします。

\(\displaystyle\int_{0}^{\pi} x f(\sin x) dx\) を簡単な形に変形します。

 

準備

\(\displaystyle\int_{0}^{\pi} x f(\sin x) dx\)

 

\(=\displaystyle\int_{\pi}^{0} (\pi-t) f(\sin (\pi-t)) (-dt)\) \(x=\pi-t\)と変換。

 

\(=\displaystyle\int_{0}^{\pi} (\pi-t) f(\sin t) dt\)

 

\(=\pi\displaystyle\int_{0}^{\pi}f(\sin t) dt-\displaystyle\int_{0}^{\pi} t f(\sin t) dt\)

 

\(=\pi\displaystyle\int_{0}^{\pi}f(\sin x) dx-\displaystyle\int_{0}^{\pi} x f(\sin x) dx\)

 

つまり   \(\displaystyle\int_{0}^{\pi} x f(\sin x) dx=\displaystyle\frac{\pi}{2}\displaystyle\int_{0}^{\pi}f(\sin x) dx\)

 

本題

今回の場合、\(f(\sin x)=\displaystyle\frac{\sin^3 x}{\sin^2 x+3}\) となる。

 

\(\displaystyle\int_{0}^{\pi} \displaystyle\frac{x\sin^3 x}{\sin^2 x+3}dx\)

 

\(=\displaystyle\frac{\pi}{2} \displaystyle\int_{0}^{\pi}\displaystyle\frac{\sin^3 x}{\sin^2 x+3}dx\) 

 

\(=\displaystyle\frac{\pi}{2} \displaystyle\int_{0}^{\pi}\displaystyle\frac{\sin x(\sin^2 x+3)-3\sin x}{\sin^2 x+3}dx\) 

 

\(=\displaystyle\frac{\pi}{2} \displaystyle\int_{0}^{\pi} \biggl(\sin x-\displaystyle\frac{3\sin x}{\sin^2 x+3}\biggr)dx\) 

 

\(=\pi-\displaystyle\frac{\pi}{2}\displaystyle\int_{1}^{-1} \displaystyle\frac{-3}{1-t^2+3}dt\)   \(t=\cos x\)

 

\(=\pi-\displaystyle\frac{3\pi}{2}\displaystyle\int_{1}^{-1} \displaystyle\frac{1}{t^2-4}dt\)

 

\(=\pi+\displaystyle\frac{3\pi}{2\times 4}\displaystyle\int_{-1}^{1} \biggl(\displaystyle\frac{1}{t-2}-\displaystyle\frac{1}{t+2}\biggr) dt\)

 

\(=\pi+\displaystyle\frac{3\pi}{8}\biggl[\log\biggl|\displaystyle\frac{t-2}{t+2}\biggr|\biggr]_{-1}^{1}\)

 

\(=\pi+\displaystyle\frac{3\pi}{8}\biggl(\log\biggl|\displaystyle\frac{1}{3}\biggr|-\log\biggl|\displaystyle\frac{-3}{1}\biggr|\biggr)\)

 

\(=\pi-\displaystyle\frac{3}{4}\pi\log 3\)

 

答え

\(\pi-\displaystyle\frac{3}{4}\pi\log 3\)

 

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