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積分問題71番
思考
対称性を利用します。（\(x\)を\(-x\)に置き換えたり）
計算
\(I=\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\displaystyle\frac{\sin ^2 x}{1+e^x}dx\)とおく。
\(x=-t\)とおくと
\(I=\displaystyle\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}\displaystyle\frac{\sin ^2 t}{1+e^{-t}}(-dt)=\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\displaystyle\frac{e^t\sin ^2 t}{e^t+1}dt\)
\(=\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\displaystyle\frac{e^x\sin ^2 x}{e^x+1}dx\)
ここで青字の二式を足すと
\(2I\)\(=\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\displaystyle\frac{\sin ^2 x}{1+e^x}dx+\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\displaystyle\frac{e^x\sin ^2 x}{1+e^x}dx\)
\(=\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin ^2 x dx\)
\(=\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\displaystyle\frac{1-\cos 2x}{2}dx\) （半角公式利用）
\(=\biggl[\displaystyle\frac{1}{2}x-\displaystyle\frac{1}{4}\sin 2x\biggr]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\displaystyle\frac{\pi}{2}\)
\(2I=\displaystyle\frac{\pi}{2}\)より\(I=\displaystyle\frac{\pi}{4}\)
答え
\(\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\displaystyle\frac{\sin ^2 x}{1+e^x}dx=\displaystyle\frac{\pi}{4}\)
