定積分 $\displaystyle \int^{\frac{\pi}{3}}_{0} \frac{dx}{\cos x}$ の値を $\displaystyle t = \tan{\frac{x}{2}}$と変数変換して求める

公式

$\displaystyle t = \tan{\frac{x}{2}}$のとき、

• $\displaystyle \frac{dx}{dt} = \frac{2}{1+t^{2}}$
• $\displaystyle \sin{x} = \frac{2t}{1+t^{2}}$
• $\displaystyle \cos{x} = \frac{1-t^{2}}{1+t^{2}}$

解き方

$\displaystyle t = \tan{\frac{x}{2}}$のとき、

$\tan{\frac{x}{2}}$	$t$
$\frac{\pi}{3}$	$\tan(\frac{1}{2}\cdot \frac{\pi}{3}) = \frac{1}{\sqrt{3}}$
$0$	$\tan({\frac{1}{2}\cdot 0}) = 0$
$\frac{1}{\cos{x}}$	$\frac{1+t^{2}}{1-t^{2}}$
$dx$	$\frac{2}{1+t^{2}}dt$

と置換できるので、

\begin{eqnarray} && \int^{\frac{\pi}{3}}_{0} \frac{dx}{\cos x} \\ &=& \int^{\frac{1}{\sqrt{3}}}_{0} \left(\frac{1}{1+t} + \frac{1}{1-t}\right) dt\\ &=& \left[ \log(1+t) + \log(1-t) \right]^{\frac{1}{\sqrt{3}}}_{0}\\ &=& \left[ \log{\frac{1+t}{1-t}} \right]^{\frac{1}{\sqrt{3}}}_{0}\\ &=& \log{\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}}\\ &=& \log{\frac{(\sqrt{3}+1)^{2}}{3-1}}\\ &=& \log{\frac{1}{2} \cdot (3 + 2\sqrt{3} + 1)} \\ &=& \log{(2 + \sqrt{3})} \end{eqnarray}