How can I integrate $$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\tan x}}dx\ \ \ ?$$
I have made the integral into the form of $\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}$, but been able to go no further.
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mathlove
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four_lines
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Just a naïve suggestion : you may try changing the variable to $u = \sqrt{\tan x}$. – Joel Cohen Jun 23 '15 at 21:49
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1This integral has an explicit solution. – Zach466920 Jun 23 '15 at 21:49
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Let $I$ be your integral.
Setting $u=\frac{\pi}{2}-x$ and using $\tan\left(\frac{\pi}{2}-x\right)=\frac{1}{\tan x}$, we have
$$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\tan x}}dx=\int_{\frac{\pi}{3}}^{\frac{\pi}{6}}\frac{-1}{1+\sqrt{\frac{1}{\tan u}}}du=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\tan u}}{1+\sqrt{\tan u}}du=J.$$ Now note that $$I+J=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}.$$ Hence, $$I=\color{red}{\frac{\pi}{12}}.$$
mathlove
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Heh. Sometimes there are ways to find a definite integral without finding an antiderivative.
Note that $\tan(\pi/2-t)=1/\tan(t)$. Say $I$ is your integral. The change of variables $t\mapsto\pi/2-t$ shows that $$I=\int_{\pi/6}^{\pi/3}\frac{1}{1+\sqrt{1/\tan(t)}}\,dt=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\tan(t)}}{1+\sqrt{\tan(t)}}\,dt.$$ Hence $$I+I=\int_{\pi/6}^{\pi/3}\,dt.$$(That's as far as I got, sorry...)
David C. Ullrich
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Heck of a feeling, you know you're almost there, can't quite get it... – David C. Ullrich Jun 23 '15 at 22:03
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Anyway ... nice solution +1 It is reminiscent of this solution, which is more general. – Mark Viola Jun 23 '15 at 22:17
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And I didn't know my comments had an invisible delete button until I looked for it after you said we could delete. So everybody learned something... – David C. Ullrich Jun 23 '15 at 22:27
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$$\begin{align}\int_{\pi/6}^{\pi/3}\frac{dx}{1 + \sqrt{\tan x}} &= \int_{\pi/6}^{\pi/4}\frac{dx}{1 + \sqrt{\tan x}} +\int_{\pi/4}^{\pi/3}\frac{dx}{1 + \sqrt{\tan x}}\\ &=\int_{\pi/6}^{\pi/4}\frac{dx}{1 + \sqrt{\tan x}} +\int_{\pi/6}^{\pi/4}\frac{\sqrt {\tan t}\, dt}{1 + \sqrt{\tan t}}, \quad t = \pi/2 -x\\ &=\int_{\pi/6}^{\pi/4} 1 \, dx\\ &= \frac{\pi}{12} \end{align}$$
abel
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