(1) \(\displaystyle\int _ 0^\pi x^2 \cos^2 x \, dx\) を求めよ.
(2) 定数 \(a\) に対して, \[ f(x) = ax \sin x +x +\dfrac{\pi}{2} \] とおく. このとき, 不等式 \[ \displaystyle\int _ 0^\pi \left\{ f'(x) \right\}^2 \, dx \geqq f \left( \dfrac{\pi}{2} \right) \] を満たす \(a\) の範囲を求めよ. ただし, \(f'(x)\) は \(f(x)\) の導関数とする.
【 解 答 】
(1)
\[\begin{align} x^2 \cos^2 x & = \dfrac{x^2}{2} +\dfrac{x^2 \cos 2x}{2} \\ & = \dfrac{\left( x^3 \right)'}{6} +\dfrac{1}{4} \left( x^2 \sin 2x \right)' -\dfrac{1}{2} x \sin 2x \\ & = \dfrac{\left( x^3 \right)'}{6} +\dfrac{1}{4} \left( x^2 \sin 2x \right)' +\dfrac{1}{4} \left( x \cos 2x \right)' +\dfrac{1}{2} \cos 2x \\ & = \left( \dfrac{x^3}{6} +\dfrac{x^2 \sin 2x}{4} +\dfrac{x \cos 2x}{4} +\dfrac{\sin 2x}{4}\right)' \end{align}\] よって \[\begin{align} \displaystyle\int _ 0^\pi x^2 \cos^2 x \, dx & = \left[ \dfrac{x^3}{6} +\dfrac{x^2 \sin 2x}{4} +\dfrac{x \cos 2x}{4} +\dfrac{\sin 2x}{4}\right] _ 0^\pi \\ & = \underline{\dfrac{{\pi}^3}{6} +\dfrac{\pi}{4}} \end{align}\]
(2)
条件より \[\begin{align} f \left( \dfrac{\pi}{2} \right) & = \pi \left( \dfrac{a}{2} +1 \right) , \\ f'(x) & = a \left( \sin x +x \cos x \right) +1 \end{align}\] なので \[\begin{align} \left\{ f'(x) \right\}^2 & = a^2 \left( \sin^2 x +2x \sin x \cos x +x^2 \cos^2 x \right) \\ & \qquad +2a \left( \sin x +x \cos x \right) +1 \\ & = a^2 \left( x \sin^2 x \right)' +2a \left( x \sin x \right)' +a^2 x^2 \cos^2 x +1 \end{align}\] したがって \[\begin{align} \displaystyle\int _ 0^\pi \left\{ f'(x) \right\}^2 \, dx & = \left[ a^2x \sin^2 x +2ax \sin x +x \right] _ 0^\pi \\ & \qquad +a^2 \left( \dfrac{{\pi}^3}{6} +\dfrac{\pi}{4} \right) \\ & = a^2 \left( \dfrac{{\pi}^3}{6} +\dfrac{\pi}{4} \right) +\pi \end{align}\] 以上より, \(\displaystyle\int _ 0^\pi \left\{ f'(x) \right\}^2 \, dx \geqq f \left( \dfrac{\pi}{2} \right)\) をとくと \[\begin{gather} a^2 \left( \dfrac{{\pi}^3}{6} +\dfrac{\pi}{4} \right) +\pi \geqq \pi \left( \dfrac{a}{2} +1 \right) \\ a \left\{ ( 2{\pi}^2 +3 ) a -6 \right\} \geqq 0 \\ \text{∴} \quad \underline{a \leqq 0 , \ \dfrac{6}{2{\pi}^2 +3} \leqq a} \end{gather}\]