[2007年埼玉医科大学・数学]定積分

\[ I=\int_{0}^{\frac{\pi}{6}}\frac{1}{1+\sqrt{3}\tan x}dx \] とおく。
分母分子に\(\cos x\)を掛けると、
\begin{align*} I & =\int_{0}^{\frac{\pi}{6}}\frac{1}{1+\sqrt{3}\tan x}dx\\ & =\int_{0}^{\frac{\pi}{6}}\frac{\cos x}{\cos x+\sqrt{3}\sin x}dx \end{align*} 分母は3角関数の合成より、
\begin{align*} \cos x+\sqrt{3}\sin x & =\sqrt{1^{2}+\left(\sqrt{3}\right)^{2}}\left(\frac{1}{\sqrt{1^{2}+\left(\sqrt{3}\right)^{2}}}\cos x+\frac{\sqrt{3}}{\sqrt{1^{2}+\left(\sqrt{3}\right)^{2}}}\sin x\right)\\ & =2\left(\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\right)\\ & =2\left(\sin\frac{\pi}{6}\cos x+\cos\frac{\pi}{6}\sin x\right)\\ & =2\sin\left(x+\frac{\pi}{6}\right) \end{align*} となるので、
\begin{align*} I & =\int_{0}^{\frac{\pi}{6}}\frac{\cos x}{\cos x+\sqrt{3}\sin x}dx\\ & =\int_{0}^{\frac{\pi}{6}}\frac{\cos x}{2\sin\left(x+\frac{\pi}{6}\right)}dx \end{align*} となる。
ここで\(t=x+\frac{\pi}{6}\)とおくと、\(x:0\rightarrow\frac{\pi}{6}\)のとき、\(t:\frac{\pi}{6}\rightarrow\frac{\pi}{3}\)となるので、
\begin{align*} I & =\int_{0}^{\frac{\pi}{6}}\frac{\cos x}{2\sin\left(x+\frac{\pi}{6}\right)}dx\\ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\cos\left(t-\frac{\pi}{6}\right)}{2\sin t}dt \end{align*} となる。
分子は加法定理より、
\begin{align*} \cos\left(t-\frac{\pi}{6}\right) & =\cos t\cos\frac{\pi}{6}+\sin t\sin\frac{\pi}{6}\\ & =\frac{\sqrt{3}}{2}\cos t+\frac{1}{2}\sin t \end{align*} となるので、
\begin{align*} I & =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\cos\left(t-\frac{\pi}{6}\right)}{2\sin t}dt\\ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\frac{\sqrt{3}}{2}\cos t+\frac{1}{2}\sin t}{2\sin t}dt\\ & =\frac{\sqrt{3}}{4}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\cos t}{\sin t}dt+\frac{1}{4}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}dt\\ & =\frac{\sqrt{3}}{4}\left[\log\left(\sin t\right)\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}+\frac{1}{4}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)\\ & =\frac{\sqrt{3}}{4}\left(\log\left(\sin\frac{\pi}{3}\right)-\log\left(\sin\frac{\pi}{6}\right)\right)+\frac{1}{4}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)\\ & =\frac{\sqrt{3}}{4}\left(\log\frac{\sqrt{3}}{2}-\log\frac{1}{2}\right)+\frac{1}{4}\cdot\frac{\pi}{6}\\ & =\frac{\sqrt{3}}{4}\log\sqrt{3}+\frac{\pi}{24}\\ & =\frac{\sqrt{3}}{8}\log3+\frac{\pi}{24} \end{align*}