(1) \(f(x)\) を連続関数とするとき, \[ \displaystyle\int _ 0^\pi x f \left( \sin x \right) dx = \dfrac{\pi}{2} \displaystyle\int _ 0^\pi f \left( \sin x \right) dx \] が成り立つことを示せ.
(2) 定積分 \[ \displaystyle\int _ 0^\pi \dfrac{x \sin^3 x}{\sin^2 x +8} \, dx \] の値を求めよ.
【 解 答 】
(1)
\(I = \displaystyle\int _ 0^\pi x f \left( \sin x \right) dx\) とおく.
\(x = \pi -t\) とおくと, \(dx = -dt\) であり
\[
\begin{array}{c|ccc} x & 0 & \rightarrow & \pi \\ \hline t & \pi & \rightarrow & 0 \end{array}
\]
したがって
\[\begin{align}
I & = -\displaystyle\int _ 0^\pi ( \pi -t ) f \left( \sin ( \pi -t ) \right) dt \\
& = \pi \displaystyle\int _ 0^\pi f \left( \sin t \right) dt -\displaystyle\int _ 0^\pi t f \left( \sin t \right) dt \quad ( \ \text{∵} \ \sin ( \pi -t ) = \sin t \ ) \\
& = \pi \displaystyle\int _ 0^\pi f \left( \sin t \right) dt -I \\
\text{∴} \quad 2I & = \pi \displaystyle\int _ 0^\pi f \left( \sin x \right) dx
\end{align}\]
よって
\[
I = \underline{\dfrac{\pi}{2} \displaystyle\int _ 0^\pi f \left( \sin x \right) dx}
\]
(2)
\(J = \displaystyle\int _ 0^\pi \dfrac{x \sin^3 x}{\sin^2 x +8} \, dx\) とおく.
\(f(x) = \dfrac{x^3}{x^2+8} = x -\dfrac{8x}{x^2+8}\) とおくと, これは連続関数なので, (1) の結果を用いて
\[\begin{align}
J & = \dfrac{\pi}{2} \displaystyle\int _ 0^\pi \left( \sin x -\dfrac{8 \sin x}{\sin ^2 x +8} \right) dx \\
& = \dfrac{\pi}{2} \displaystyle\int _ 0^\pi \sin x dx -4 \pi \displaystyle\int _ 0^\pi \dfrac{\sin x}{9 -\cos^2 x} \, dx
\end{align}\]
ここで
\[\begin{align}
\displaystyle\int _ 0^\pi \sin x \, dx & = \left[ -\cos x \right] _ 0^{\pi} = 2 , \\
\displaystyle\int _ 0^\pi \dfrac{\sin x}{9 -\cos^2 x} \, dx & = -\displaystyle\int _ 0^{\pi} \dfrac{\left( \cos x \right)'}{( 3 +\cos x )( 3 -\cos x )} \, dx \\
& = -\dfrac{1}{6} \displaystyle\int _ 0^{\pi} \left( \dfrac{1}{3 +\cos x} + \dfrac{1}{3 -\cos x} \right) \left( \cos x \right)' \, dx \\
& = -\dfrac{1}{6} \left[ \log \dfrac{3 +\cos x}{3 -\cos x} \right] _ 0^{\pi} \\
& = -\dfrac{1}{6} \left( \log \dfrac{1}{2} -\log 2 \right) \\
& = \dfrac{\log 2}{3}
\end{align}\]
したがって
\[\begin{align}
J & = \dfrac{\pi}{2} \cdot 2 -4 \pi \cdot \dfrac{\log 2}{3} \\
& = \underline{\pi \left( 1 -\dfrac{4 \log 2}{3} \right)}
\end{align}\]