You've probably seen this integral before if you paid attention

First we take a look at $N$ . Since the shortest side in a triangle is opposite the smallest angle, an easy comparison (squaring and adding/subtracting appropriately) tells us $\sqrt{41-8\sqrt{10}}$ is the shortest. Thus, by the cosine law,

$41-8\sqrt{10} = 4^2 + 5^2 - 2(4)(5)\cos N$

$\Rightarrow\cos N = \dfrac{\sqrt{2}}{\sqrt{5}}$

Now, simplifying first the indefinite integral with the pythagorean identity for $\sec$ and $\tan$ , we get

$\displaystyle \int \dfrac{\sec^2\theta}{(\sec^2\theta-\tan^2\theta)(\sec^2\theta + \tan^2\theta)} (\sec^2\theta\;d\theta)$

Using the substitution $u=\tan \theta$ , we know that $\theta=0 \Rightarrow u=0; \theta = N \Rightarrow u = \tan N = \sqrt{3}/\sqrt{2}$ . Hence,

$\Rightarrow I =\displaystyle \int_0^{\sqrt{3}/\sqrt{2}}\dfrac{1+u^2}{1+2u^2}\;du$

$=\displaystyle \dfrac{1}{2}\int_0^{\sqrt{3}/\sqrt{2}} 1 + \dfrac{1}{1+2u^2}\;du$

$= \dfrac{\pi\sqrt{2} + 3\sqrt{6}}{12}$

where the last integral simplifies with $\tan^{-1}$ .

Hence, $\alpha + \beta + \gamma + \delta = 23.$

The integrand can be simplified to 1/[(cosx)^2 (1+(sinx)^2)] after changing (theta) for x, for simplicity. Then, change the variable by letting tanx=t. After simplifying & integrating, the integral, I=1/2 tan(x)+1/(2 sqrt(2)) arctan(sqrt(2)*tan(x)). Now you can find N in radians, which is the smallest of the angles of the triangle with the above given sides. Using the Law of Cosines, N=0.886077 radians or 50.76847 degrees. Substitute this value in I & simplify according to the format of the given answer, then one could find alpha=2, beta=3, gamma=6 & phi=12, from which the sum is 23.