上級者
数学Ⅲ：積分法
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ヒント：この問題はノーヒントだとおそらく解くことは厳しい．解き方を覚えておきましょう．
問題
定積分\(I=\displaystyle \int_{0}^{\pi}\frac{x\sin^{2}x}{\sin^{6}x+\cos^{6}x}dx\)の値を求めよ．なお，\(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{\sin^{6}x+\cos^{6}x}dx\)\(=\pi\)であることを使ってもよい．（オリジナル）
解説
ここでは以下の3つの方法を覚えておこう．
(A) まずは，\(\displaystyle \int_{0}^{\pi}xf(\sin x)dx\)\(=\displaystyle \frac{\pi}{2} \int_{0}^{\pi}f(\sin x)dx\)が成り立つことを示す．
\(x=\pi-t\)と置くと，\(\sin x=\sin (\pi-t)=\sin t\)，
\(x:0\rightarrow \pi\)で\(t:\pi\rightarrow 0\)，\(dx=-dt\)
よって，\(\displaystyle \int_{0}^{\pi}xf(\sin x)dx\)\(=\displaystyle \int_{\pi}^{0}(\pi-t)f(\sin t)(-dt)\)\(=\displaystyle \int_{0}^{\pi}\pi f(\sin t)dt - \int_{0}^{\pi}tf(\sin t)dt\)
ここで，\(t\)や\(x\)は単なる積分変数なので変更可能．
よって，\(\displaystyle \int_{0}^{\pi}xf(\sin x)dx\)\(=\displaystyle \int_{0}^{\pi}\pi f(\sin x)dx - \int_{0}^{\pi}xf(\sin x)dx\) \(\Leftrightarrow\) \(\displaystyle \int_{0}^{\pi}xf(\sin x)dx\)\(=\displaystyle \frac{\pi}{2}\int_{0}^{\pi} f(\sin x)dx\)
(B)次に，\(\displaystyle \int_{0}^{\pi}\frac{\sin^{2}x}{\sin^{6}x+\cos^{6}x}dx\)\(=2\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x}{\sin^{6}x+\cos^{6}x}dx\)が成り立つことを示す．（\(x=\frac{\pi}{2}\)で左右対称となる．）
\(J=\displaystyle \int_{\frac{\pi}{2}}^{\pi}\frac{\sin^{2}x}{\sin^{6}x+\cos^{6}x}dx\)に関して，\(x=\pi-t\)と置くと，
\(\sin x=\sin (\pi-t)=\sin t\)，\(\cos x=\cos (\pi-t)=-\cos t\)
\(\displaystyle x:\frac{\pi}{2}\rightarrow \pi\)で\(\displaystyle t:\frac{\pi}{2}\rightarrow 0\)，\(dx=-dt\)
よって，\(J=\displaystyle \int_{\frac{\pi}{2}}^{0}\frac{\sin^{2}t}{\sin^{6}t+\cos^{6}t}(-dt)\)\(=\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x}{\sin^{6}x+\cos^{6}x}dx\) (\(t\)や\(x\)は単なる積分変数なので変更可能)
よって，\(\displaystyle \int_{0}^{\pi}\frac{\sin^{2}x}{\sin^{6}x+\cos^{6}x}dx\)\(=2\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x}{\sin^{6}x+\cos^{6}x}dx\)が成り立つ．
(C)次に，\(\displaystyle \int_{0}^{\frac{\pi}{2}}f(\sin x)dx\)\(=\displaystyle \int_{0}^{\frac{\pi}{2}}f(\cos x)dx\)が成り立つことを示す．
\(\displaystyle x=\frac{\pi}{2}-t\)と置くと，\(\sin x\)\(=\displaystyle \sin \left(\frac{\pi}{2}-t\right)\)\(=\cos t\)，また\(\cos x\)\(=\displaystyle \sin \left(\frac{\pi}{2}-t\right)\)\(=\sin t\)
\(\displaystyle x:0\rightarrow \frac{\pi}{2}\)で\(\displaystyle t:\frac{\pi}{2}\rightarrow 0\)，\(dx=-dt\)
よって，\(\displaystyle \int_{0}^{\frac{\pi}{2}}f(\sin x)dx\)\(=\displaystyle \int_{\frac{\pi}{2}}^{0}f(\cos t)(-dt)\)\(=\displaystyle \int_{0}^{\frac{\pi}{2}}f(\cos x)dx\)が成り立つ．(\(t\)や\(x\)は単なる積分変数なので変更可能)
以上の性質を利用すると，\(I=\displaystyle \int_{0}^{\pi}\frac{x\sin^{2}x}{\sin^{6}x+\cos^{6}x}dx\)\(=\displaystyle \frac{\pi}{2}\int_{0}^{\pi}\frac{\sin^{2}x}{\sin^{6}x+\cos^{6}x}dx\)\(=\pi\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x}{\sin^{6}x+\cos^{6}x}dx\)\(=\pi\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos^{2}x}{\sin^{6}x+\cos^{6}x}dx\)
よって，\(2I=\pi\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x}{\sin^{6}x+\cos^{6}x}dx\)\(+\pi\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos^{2}x}{\sin^{6}x+\cos^{6}x}dx\)\(=\pi\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{\sin^{6}x+\cos^{6}x}dx\)
ここで，\(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{\sin^{6}x+\cos^{6}x}dx\)\(=\pi\)を代入して整理すれば，\(\displaystyle I=\frac{\pi^2}{2}\)となる．
(1) \(f(x)\)を連続関数とするとき，\(\displaystyle \int_{0}^{\pi}xf(\sin x)dx\)\(=\displaystyle \frac{\pi}{2}\int_{0}^{\pi}f(\sin x)dx\)が成り立つことを示せ．
(2) 定積分\(\displaystyle \int_{0}^{\pi}\frac{x\sin^{3}x}{\sin^{2}x +8}dx\)の値を求めよ．（2010 横浜国立大）
(1)については，上記の(A)に記載．
(2)の答えは，\(\displaystyle \left(1-\frac{4\log 2}{3}\right)\pi\)