工学部2020年第１問（１）

工学部

2020.02.212020.07.18

問題文全文
着眼点
解答

問題文全文

　以下の定積分の値を求めなさい. ただし, $\log x$ は $x$ の自然対数を表す.

(a) $\displaystyle \int_0^{\frac{\pi}{4}}\frac{\sin x-\sqrt{2}\cos x}{\sqrt{2}\sin x+\cos x}dx=\log\left(\fbox{$\hskip0.8emア\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}-\sqrt{\fbox{$\hskip0.8emイ\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}}\right)$

(b) $\displaystyle \int_0^{\frac{\pi}{4}}\frac{\sin x}{\sqrt{2}\sin x+\cos x}dx=\frac{\sqrt{\fbox{$\hskip0.8emウ\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}}}{\fbox{$\hskip0.8emエオ\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}}\pi +\frac{\fbox{$\hskip0.8emカ\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}}{\fbox{$\hskip0.8emキ\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}}\log \left(\fbox{$\hskip0.8emク\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}-\sqrt{\fbox{$\hskip0.8emケ\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}}\right)$

(c) $\displaystyle \int_0^{\frac{\pi}{4}}\frac{\cos x}{\sqrt{2}\sin x+\cos x}dx=\frac{\fbox{$\hskip0.8emコ\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}}{\fbox{$\hskip0.8emサシ\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}}\pi -\frac{\sqrt{\fbox{$\hskip0.8emス\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}}}{\fbox{$\hskip0.8emセ\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}}\log \left(\fbox{$\hskip0.8emソ\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}-\sqrt{\fbox{$\hskip0.8emタ\hskip0.8em\Rule{0pt}{0.8em}{0.4em}$}}\right)$

着眼点

(a)は明らかに $\displaystyle \int \frac{f^{\prime}(x)}{f(x)}dx=\log f(x)+C$ を利用する形です.

(b),(c) は被積分関数の分母が同じですし, 分母も $\sin x$ か $\cos x$ かの違いだけです. 確実に関係があると見ていいでしょう. 簡単のために (b) の式を $\mathrm{B}$ , (c) の式を $\mathrm{C}$ とおくと, (a) の式 $\mathrm{A}$ が $\mathrm{B}$ と $\mathrm{C}$ で表せます. もう一つ $\mathrm{B}$ と $\mathrm{C}$ の関係式が作れれば解くことができそうです.

解答

(a)

$\begin{align} \int_0^{\frac{\pi}{4}}\frac{\sin x-\sqrt{2}\cos x}{\sqrt{2}\sin x+\cos x}dx=-\int_0^{\frac{\pi}{4}}\frac{\left(\sqrt{2}\sin x+\cos x\right)^{\prime}}{\sqrt{2}\sin x+\cos x}dx\end{align}$

$\begin{align}=\biggl[-\log \left(\sqrt{2}\sin x+\cos x\right)\biggr]_0^{\frac{\pi}{4}}=-\log \left(1+\frac{1}{\sqrt{2}}\right)\end{align}$

$\begin{align}=\log \frac{\sqrt{2}}{\sqrt{2}+1}=\log \left(2-\sqrt{2}\right)\end{align}$

(b), (c)

$\begin{align}B=\int_0^{\frac{\pi}{4}}\frac{\sin x}{\sqrt{2}\sin x+\cos x}dx, C=\int_0^{\frac{\pi}{4}}\frac{\cos x}{\sqrt{2}\sin x+\cos x}dx\end{align}$

とおくと, (1) より

$\begin{align} B-\sqrt{2}C=\log \left(2-\sqrt{2}\right)\cdots ①\end{align}$

が成り立つ.

一方,

$\begin{align}\sqrt{2}B+C=\int_0^{\frac{\pi}{4}}\frac{\sqrt{2}\sin x+\cos x}{\sqrt{2}\sin x+\cos x}dx=\int_0^{\frac{\pi}{4}}dx=\frac{\pi}{4}\cdots ②\end{align}$

が成り立つ. ①, ②より