京都大学(理系) 2025年　問題1-2


$次の定積分の値を求めよ。$
\[(1)\ \ \int_0^{\sqrt{3}}\dfrac{x\sqrt{x^2+1}+2x^3+1}{x^2+1} dx\] \[(2)\ \ \int_0^{\scriptsize{\dfrac{\pi}{2}}}\sqrt{\dfrac{1-\cos x}{1+\cos x}} dx\]



(1)


\begin{eqnarray*} I &=&\int_0^{\sqrt{3}}\dfrac{x\sqrt{x^2+1}+2x^3+1}{x^2+1} dx\\ \\ &=&\int_0^{\sqrt{3}}\big(\dfrac{x}{\sqrt{x^2+1}} + \dfrac{2x(x^2+1)-2x+1}{x^2+1}\big) dx\\ \\ &=&\int_0^{\sqrt{3}}\big(\dfrac{x}{\sqrt{x^2+1}} + 2x -\dfrac{2x}{x^2+1}+\dfrac{1}{x^2+1}\big) dx\\ \end{eqnarray*}
\begin{eqnarray*} I_1 &=&\int_0^{\sqrt{3}}\big(\dfrac{x}{\sqrt{x^2+1}} + 2x -\dfrac{2x}{x^2+1}\big) dx\\ \\ &=&\big[\sqrt{x^2+1} +x^2-\log(x^2+1)\big]_0^{\sqrt{3}}\\ \\ &=&2+3-\log 4-1\\ \\ &=&4-2\log 2 \end{eqnarray*}
\[I_2=\int_0^{\sqrt{3}}\dfrac{1}{x^2+1} dx \quad は \] \[ x=\tan \theta \quad とおくと \quad dx=\dfrac{d\theta}{\cos ^2\theta} \qquad \begin{array}{c|c} x & \ 0\ \rightarrow \sqrt{3} \quad \\ \hline \theta & 0\ \rightarrow \dfrac{\pi}{3} \\ \end{array} \]
\begin{eqnarray*} I_2 &=&\int_0^{\scriptsize{\dfrac{\pi}{3}}} \dfrac{\dfrac{d\theta}{\cos ^2\theta}}{\tan^2 \theta +1}\\ \\ &=&\int_0^{\scriptsize{\dfrac{\pi}{3}}} \dfrac{d\theta}{\cos ^2\theta(1+\tan^2 \theta)}\\ \\ &=&\int_0^{\scriptsize{\dfrac{\pi}{3}}} d\theta \\ \\ &=&\dfrac{\pi}{3} \end{eqnarray*} $\therefore \ \ I=I_1+I_2=4-2\log 2+\dfrac{\pi}{3}$


(2)


\begin{eqnarray*} & &\int_0^{\scriptsize{\dfrac{\pi}{2}}}\sqrt{\dfrac{1-\cos x}{1+\cos x}} dx\\ \\ &=&\int_0^{\scriptsize{\dfrac{\pi}{2}}}\sqrt{\dfrac{1-(1-2\sin^2 \scriptsize{\dfrac{x}{2}})}{1+(2\cos^2\scriptsize{\dfrac{x}{2}}-\normalsize{1})}} dx\\ \\ &=&\int_0^{\scriptsize{\dfrac{\pi}{2}}}\sqrt{\dfrac{\sin^2 \scriptsize{\dfrac{x}{2}}}{\cos^2\scriptsize{\dfrac{x}{2}}}} dx\\ \\ &=&\int_0^{\scriptsize{\dfrac{\pi}{2}}}\dfrac{\sin \scriptsize{\dfrac{x}{2}}}{\cos \scriptsize{\dfrac{x}{2}}} dx\\ \\ &=&\big[-2\log \cos \scriptsize{\dfrac{x}{2}}\big]_0^{\scriptsize{\dfrac{\pi}{2}}} \\ \\ &=&-2\big(\log \cos \dfrac{\pi}{4} - \log \cos 0\big)\\ \\ &=&-2\big(\log \dfrac{1}{\sqrt{2}} - \log 1\big)\\ \\ &=&2\log \sqrt{2}\\ \\ &=&\log 2 \end{eqnarray*}

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