The integral is from my MIT Integration Bee 2025 Mock Training Problems. But unfortunately, I forgot the solutions nor the trick to this problem.
I tried letting $u=\sqrt{x-1}$, then perform trig-substitution on $u=\sqrt{2}\sec\theta$.
However, this resulted to a nasty rational trig-function.
I tried letting $u=\sqrt{x+1}$, but then $u=\sqrt{2}\tan\theta$ results the same nasty rational trig-function.
I thought about maybe finding a sneaky way to conjugate the denominator, but that square is out of the way.
Any sneaky way to solve this integral?
Best Answer
Not an entirely satisfying answer, but at least I can show how to get an exponentially convergent series representation that evaluates to the hypergeometric result in the comment by @Mariusz Iwaniuk.
With $x=1+2\tan(x)^2$, the integral becomes $$I=4\int_{0}^{\pi/2} \frac{2\cos(x) + \sqrt{2}}{4 - \sqrt{2} \sin(x) \cos(x)^3} \sin(x)\,\text{d}x.$$ Expanding in a geometric series gives $$\sum_{n\ge 0} \int_{0}^{\pi/2} \bigl(2\cos(x) + \sqrt{2}\bigr) \sin(x) \left(\frac{\sqrt{2}}{4} \sin(x) \cos(x)^3\right)^n\,\text{d}x.$$ With the Beta function $B(x,y) = 2\int_0^{\pi/2} \sin(t)^{2x-1} \cos(t)^{2y-1}\,\text{d}t$ we get $$\sum_{n\ge 0} \left(\frac{\sqrt{2}}{4}\right)^n \left\{ B\left(1+\frac{n}{2}, 1 + \frac{3n}{2}\right) + B\left(1+\frac{n}{2}, \frac{1+3n}{2}\right) \right\}.$$ With $B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ this evaluates to $$I = \sum_{n\ge 0} 2^{-3n/2} \Gamma\left(1 + \frac{n}{2}\right) \left\{ \frac{\Gamma\left(1 + \frac{3n}{2}\right)}{\Gamma\left(2 + 2n\right)} + \frac{\Gamma\left(\frac{1+3n}{2}\right)}{\Gamma\left(\frac{3}{2} + 2n\right)} \right\}.$$ Using WolframAlpha, the even $n$ sum up to $${}_4F_3\left( \frac{1}{3}, \frac{2}{3}, 1, 1; \frac{2}{4}, \frac{3}{4}, \frac{5}{4}; \frac{27}{2048} \right) + \sqrt{2} {}_5F_4\left( \frac{1}{6}, \frac{3}{6}, \frac{5}{6}, 1, 1; \frac{3}{8}, \frac{5}{8}, \frac{7}{8}, \frac{9}{8}; \frac{27}{2048} \right)$$ and the odd $n$ sum to $$\frac{1}{15} {}_5F_4\left( \frac{2}{3}, 1, 1, \frac{4}{3}, \frac{3}{2}; \frac{7}{8}, \frac{9}{8}, \frac{11}{8}, \frac{13}{8}; \frac{27}{2048} \right) + \frac{\pi}{32\sqrt{2}} {}_3F_2\left( \frac{5}{6}, \frac{7}{6}, \frac{9}{6}; \frac{5}{4}, \frac{7}{4}; \frac{27}{2048} \right).$$