# Evaluation of $\int_0^1\arctan t^2 + \frac\pi4\sqrt{\tan\left(\frac\pi 4t\right)}\ \text{dt}$

$$\int_0^1\arctan t^2 + \frac\pi4\sqrt{\tan\left(\frac\pi 4t\right)}\ \text{dt}$$ I've found that individually, $\arctan t^2$ and $\sqrt{\tan t}$ are not easily integrated individually. Wolfram alpha has no closed-form solution, and the general integration involves non-elementary functions.

If this question is not answered within 48 hours I intend to self-answer.

• So this is kind of a challenge question for us. Right? – Jaideep Khare Jun 6 '17 at 1:46
• The first part of the integral is easily evaluated using integration by parts. Possibly the second as well. – Simply Beautiful Art Jun 6 '17 at 1:51
• For, $\int\arctan(t^2)dt$, we can get a closed form (although not verypretty): $$\dfrac{2^\frac{3}{2}x\arctan\left(x^2\right)+\ln\left(x\left(x+\sqrt{2}\right)+1\right)-\ln\left(x\left(x-\sqrt{2}\right)+1\right)-2\left(\arctan\left(\sqrt{2}x+1\right)+\arctan\left(\sqrt{2}x-1\right)\right)}{2^\frac{3}{2}}+C$$and we can also get a closed form for the other integral. Overall result is $\frac{\pi}{4}$ after evaluation. There is probably an easier method though I'm guessing. – user12345 Jun 6 '17 at 1:52
• This looks a bit like one of those funny ones with $f(x)+f^{-1}(x)$. But $\sqrt{\tan{x}}$ actually does has an elementary antiderivative. – Chappers Jun 6 '17 at 1:53
• @Chappers GAH! I should have seen that sooner! Oh well... its too late in the night for this anyways, so I wish everyone good integration. – Simply Beautiful Art Jun 6 '17 at 1:54

Note that $$\int_0^1 {\frac{\pi }{4}\sqrt {\tan \left( {\frac{\pi }{4}x} \right)} dx} = \int_0^{\frac{\pi }{4}} {\sqrt {\tan x} dx}$$
Therefore original integral is $$\int_0^1 {\arctan {y^2}dy} + \int_0^{\frac{\pi }{4}} {\sqrt {\tan x} dx}$$
We think about the graph of $y=\sqrt{\tan x}$, we see that the sum is the area of a rectangle with side lengths $1$ and $\pi/4$. So the result is $\pi/4$