神戸大学(理系) 2021年 問題2
$次の定積分を求めよ。$
\[(1)\ \ I=\int _0^1x^2\sqrt{1-x^2}dx\]
\[(2)\ \ J=\int _0^1x^3\log(x^2+1)dx\]
$(解説)$
$(1)も(2)も被積分関数をうまく変形して、適当に置換積分します。$
(1)
\begin{eqnarray*} I &=&\int _0^1 x^2\sqrt{1-x^2}dx\\ &=&\int _0^1 \{-(1-x^2-1)\}\sqrt{1-x^2}dx\\ &=&\int _0^1 \big\{-(1-x^2)^{\scriptsize{\cfrac{3}{2}}} + \sqrt{1-x^2}\big\}dx\\ \end{eqnarray*} \[\quad x=\sin \theta \quad とおくと \quad dx=\cos \theta d\theta \hspace{3em} \begin{array}{c|c} x & 0\ \ \rightarrow \ \ 1 \\ \hline \theta & 0\ \ \rightarrow \ \cfrac{\pi}{2} \\ \end{array} \] $\hspace{3em} 1-x^2=1-\sin ^2 \theta=\cos ^2 \theta $
\begin{eqnarray*} I &=&\int _0^{\scriptsize{\cfrac{\pi}{2}}}(-\cos ^3 \theta + \cos \theta )\cos \theta d\theta\\ &=&\int _0^{\scriptsize{\cfrac{\pi}{2}}}(1-\cos ^2 \theta ) \cos ^2 \theta d\theta\\ &=&\int _0^{\scriptsize{\cfrac{\pi}{2}}}\sin ^2 \theta \cos ^2 \theta d\theta\\ &=&\int _0^{\scriptsize{\cfrac{\pi}{2}}}\Big(\cfrac{\sin 2 \theta}{2}\Big)^2 d\theta\\ &=&\cfrac{1}{4}\int _0^{\scriptsize{\cfrac{\pi}{2}}}\cfrac{1-\cos 4 \theta}{2} d\theta\\ &=&\cfrac{1}{8}\big[\theta - \cfrac{1}{4}\sin 4\theta \ \big]_0^{\scriptsize{\cfrac{\pi}{2}}}\\ &=&\cfrac{1}{8}\big(\cfrac{\pi}{2}- \cfrac{1}{4}\sin 2\pi \big)\\ &=&\cfrac{\pi}{16}\\ \end{eqnarray*}
(2)
\[J=\int _0^1x^3\log(x^2+1)dx=\int _0^1x^2\log(x^2+1)\cdot xdx\]
\[\quad x^2+1=t \quad とおくと \quad 2xdx=dt \qquad \begin{array}{c|c} x & 0\ \ \rightarrow \ \ 1\\ \hline t & 1\ \ \rightarrow \ \ 2\\ \end{array} \]
\begin{eqnarray*} J &=&\int _1^2 (t-1)\log t \cdot \cfrac{dt}{2}\\ \\ &=&\cfrac{1}{2}\Big[(\cfrac{t^2}{2}-t)\log t \Big]_1^2-\cfrac{1}{2}\int _1^2 (\cfrac{t^2}{2}-t) \cfrac{1}{t}dt\\ \\ &=&-\cfrac{1}{2}\int _1^2 \big(\cfrac{t}{2}-1 \big)dt\\ \\ &=&\cfrac{1}{8}\\ \end{eqnarray*}
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