群馬大学(医・電子・機械) 2025年 問題3
$次の定積分を求めよ。$
\[(1)\ \ \int_0^1 \dfrac{x}{\sqrt{x+1}}dx \hspace{5em} (2)\ \ \int_0^1 \dfrac{x^3}{\sqrt{2-x^2}}dx \hspace{5em} (3)\ \ \int_0^1 \dfrac{x^2}{(1+x^2)\sqrt{1+x^2}}dx\]
(1)
\begin{eqnarray*} & &\int_0^1 \dfrac{x}{\sqrt{x+1}}dx\\ \\ &=&\int_0^1 \dfrac{(x+1)-1}{\sqrt{x+1}}dx\\ \\ &=&\int_0^1 \big(\sqrt{x+1}- \dfrac{1}{\sqrt{x+1}}\big)dx\\ \\ &=&\int_0^1 \big((x+1)^{\scriptsize{\dfrac{1}{2}}}- (x+1)^{-\scriptsize{\dfrac{1}{2}}}\big)dx\\ \\ &=&\big[\dfrac{2}{3}(x+1)^{\scriptsize{\dfrac{3}{2}}}- 2(x+1)^{\scriptsize{\dfrac{1}{2}}}\big]_0^1\\ \\ &=&\dfrac{2}{3} \times 2^{\scriptsize{\dfrac{3}{2}}}- 2 \times 2^{\scriptsize{\dfrac{1}{2}}} -(\dfrac{2}{3}-2)\\ \\ &=&\dfrac{4}{3}\sqrt{2} -2\sqrt{2} +\dfrac{4}{3}\\ \\ &=&\dfrac{4-2\sqrt{2}}{3} \end{eqnarray*}
(2)
\[I=\int_0^1 \dfrac{x^3}{\sqrt{2-x^2}}dx=\int_0^1 \dfrac{x^2 \cdot x }{\sqrt{2-x^2}}dx\] \[ 2-x^2=t \quad とおくと \quad -2xdx=dt \qquad \begin{array}{c|c} x & 0\ \ \rightarrow 1 \quad \\ \hline t & 2\ \ \rightarrow 1 \\ \end{array} \] \begin{eqnarray*} I &=&\int _2^1\dfrac{(2-t) \times \big(-\dfrac{dt}{2}\big)}{\sqrt{t}}\\ \\ &=&\dfrac{1}{2} \int _2^1\dfrac{t-2}{\sqrt{t}}dt\\ \\ &=&\dfrac{1}{2} \int_2^1 \big( t^{\scriptsize{\dfrac{1}{2}}}- 2t^{-\scriptsize{\dfrac{1}{2}}}\big)dx\\ \\ &=&\dfrac{1}{2}\big[\dfrac{2}{3}t^{\scriptsize{\dfrac{3}{2}}}- 2 \times 2t^{\scriptsize{\dfrac{1}{2}}}\big]_2^1\\ \\ &=&\big[\dfrac{1}{3}t^{\scriptsize{\dfrac{3}{2}}}- 2t^{\scriptsize{\dfrac{1}{2}}}\big]_2^1\\ \\ &=&\big(\dfrac{1}{3} -2\big) - \big(\dfrac{1}{3} \times 2^{\scriptsize{\dfrac{3}{2}}} - 2 \times 2^{\scriptsize{\dfrac{1}{2}}}\big)\\ \\ &=&-\dfrac{5}{3}+\dfrac{4}{3}\sqrt{2}\\ \\ &=&\dfrac{-5+4\sqrt{2}}{3} \end{eqnarray*}
(3)
\[I=\int_0^1 \dfrac{x^2}{(1+x^2)\sqrt{1+x^2}}dx\] \[ x=\tan \theta \quad とおくと \quad dx=\dfrac{d\theta}{\cos ^2\theta} \qquad \begin{array}{c|c} x & 0\ \ \rightarrow 1 \quad \\ \hline \theta & 0\ \ \rightarrow \dfrac{\pi}{4} \\ \end{array} \] \begin{eqnarray*} I &=&\int_0^{\scriptsize{\dfrac{\pi}{4}}} \dfrac{\tan^2 \theta}{(1+\tan^2\theta)^{\scriptsize{\dfrac{3}{2}}}} \cdot \dfrac{d\theta}{\cos^2\theta}\\ \\ &=&\int_0^{\scriptsize{\dfrac{\pi}{4}}} \tan^2 \theta \times \cos^3\theta \times \dfrac{d\theta}{\cos^2\theta}\\ \\ &=&\int_0^{\scriptsize{\dfrac{\pi}{4}}} \dfrac{\sin^2 \theta}{\cos^2\theta} \times \cos \theta d\theta \\ \\ &=&\int_0^{\scriptsize{\dfrac{\pi}{4}}} \dfrac{\sin^2 \theta}{1-\sin^2\theta} \times \cos \theta d\theta \\ \end{eqnarray*}
\[ \sin \theta =t \quad とおくと \quad \cos \theta d\theta=dt \qquad \begin{array}{c|c} \theta & 0\ \ \rightarrow \dfrac{\pi}{4}\\ \hline t & 0\ \ \rightarrow \dfrac{1}{\sqrt{2}} \\ \end{array} \] \begin{eqnarray*} I &=&\int_0^{\scriptsize{\dfrac{1}{\sqrt{2}}}} \dfrac{t^2}{1-t^2}dt\\ \\ &=&\int_0^{\scriptsize{\dfrac{1}{\sqrt{2}}}} \dfrac{-(1-t^2)+1}{1-t^2}dt\\ \\ &=&\int_0^{\scriptsize{\dfrac{1}{\sqrt{2}}}} \big(-1+ \dfrac{1}{1-t^2}\big)dt\\ \\ &=&\dfrac{1}{2}\int_0^{\scriptsize{\dfrac{1}{\sqrt{2}}}} \big(-2+ \dfrac{1}{1+t} + \dfrac{1}{1-t}\big)dt\\ \\ &=&\dfrac{1}{2} \big[-2t + \log(1+t) - \log(1-t)\big] _0^{\scriptsize{\dfrac{1}{\sqrt{2}}}} \\ \\ &=&\dfrac{1}{2} \big(-\dfrac{2}{\sqrt{2}} + \log \dfrac{1+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}\big) \\ \\ &=&\dfrac{1}{2} \big(-\dfrac{2}{\sqrt{2}} + \log \dfrac{\sqrt{2}+1}{\sqrt{2} -1}\big) \\ \\ &=&\dfrac{1}{2} \big(-\dfrac{2}{\sqrt{2}} + \log (\sqrt{2}+1)^2\big) \\ \\ &=&-\dfrac{\sqrt{2}}{2} + \log (\sqrt{2}+1)\\ \end{eqnarray*}
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