分母分子に3角関数を含む定積分

分母分子に3角関数を含む定積分
次の積分を求めよ。
\[ \int_{0}^{\frac{\pi}{2}}\frac{\sqrt[3]{\tan x}}{\left(\sin x+\cos x\right)^{2}}dx=? \]
\begin{align*} \int_{0}^{\frac{\pi}{2}}\frac{\sqrt[3]{\tan x}}{\left(\sin x+\cos x\right)^{2}}dx & =\int_{0}^{\frac{\pi}{2}}\frac{\sqrt[3]{\tan x}}{\cos^{2}x\left(\tan x+1\right)^{2}}dx\\ & =\int_{0}^{\frac{\pi}{2}}\frac{\sqrt[3]{\tan x}}{\cos^{2}x\left(\tan x+1\right)^{2}}dx\\ & =\int_{0}^{\infty}\frac{\sqrt[3]{\tan x}}{\left(\tan x+1\right)^{2}}d\tan x\\ & =B\left(1+\frac{1}{3},2-\frac{1}{3}-1\right)\\ & =B\left(\frac{4}{3},\frac{2}{3}\right)\\ & =\frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{4}{3}+\frac{2}{3}\right)}\\ & =\frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(2\right)}\\ & =\frac{1}{3}\Gamma\left(\frac{1}{3}\right)\Gamma\left(1-\frac{1}{3}\right)\\ & =\frac{1}{3}\frac{\pi}{\sin\left(\frac{1}{3}\pi\right)}\\ & =\frac{2}{3\sqrt{3}}\pi\\ & =\frac{2\sqrt{3}}{9}\pi \end{align*}

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分母分子に3角関数を含む定積分
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