次の定積分を求めよ.
(1) \(\displaystyle\int _ 0^{\frac{\pi}{4}} \dfrac{dx}{1 +\sin x}\)
(2) \(\displaystyle\int _ {\frac{4}{3}}^{2} \dfrac{dx}{x^2 \sqrt{x-1}}\)
【 解 答 】
(1)
\[\begin{align} \dfrac{1}{1 +\sin x} & = \dfrac{1 -\sin x}{\cos^2 x} \\ & = \left( \tan x +\dfrac{1}{\cos x} \right)' \end{align}\] よって \[\begin{align} \displaystyle\int _ 0^{\frac{\pi}{4}} \dfrac{dx}{1 +\sin x} & = \left[ \tan x +\dfrac{1}{\cos x} \right] _ 0^{\frac{\pi}{4}} \\ & = 1 +\sqrt{2} -1 \\ & = \underline{\sqrt{2}} \end{align}\]
(2)
求める積分値を \(I\) とおく.
\(t = \sqrt{x-1}\) とおくと
\[\begin{align}
x & = t^2 +1 \\
\text{∴} \quad dx & = 2t \, dt
\end{align}\]
また
\[
\begin{array}{c|ccc} x & \dfrac{4}{3} & \rightarrow & 2 \\ \hline t & \dfrac{1}{\sqrt{3}} & \rightarrow & 1 \end{array}
\]
なので
\[\begin{align}
I & = \displaystyle\int _ \frac{1}{\sqrt{3}}^1 \dfrac{2t}{(t^2+1)^2 t} \, dt \\
& = 2 \displaystyle\int _ \frac{1}{\sqrt{3}}^1 \dfrac{dt}{(t^2+1)^2}
\end{align}\]
さらに, \(t = \tan \theta \ \left( -\dfrac{\pi}{2} \lt \theta \lt \dfrac{\pi}{2} \right)\) とおくと
\[
dt = \dfrac{d \theta}{\cos^2 \theta}
\]
また
\[
\begin{array}{c|ccc} t & \dfrac{1}{\sqrt{3}} & \rightarrow & 1 \\ \hline \theta & \dfrac{\pi}{6} & \rightarrow & \dfrac{\pi}{4} \end{array}
\]
よって
\[\begin{align}
I & = 2 \displaystyle\int _ {\frac{\pi}{6}}^{\frac{\pi}{4}} \cos^2 \theta \, d \theta \\
& = \displaystyle\int _ {\frac{\pi}{6}}^{\frac{\pi}{4}} \left( 1 +\cos 2 \theta \right) \, d \theta \\
& = \left[ \theta +\dfrac{\sin 2 \theta}{2} \right] _ {\frac{\pi}{6}}^{\frac{\pi}{4}} \\
& = \left( \dfrac{\pi}{4} +\dfrac{1}{2} \right) -\left( \dfrac{\pi}{6} +\dfrac{\sqrt{3}}{4} \right) \\
& = \underline{\dfrac{\pi}{12} +\dfrac{2 -\sqrt{3}}{4}}
\end{align}\]