Consider the volume bounded in the first octant by the inverted cone $z = 1 - \sqrt{x^2 + y^2}$ and the plane $x + y + z = 1.$ By definition, we may compute this volume as a triple integral $$\iiint_\mathcal S 1 \, dV$$ for the three-dimensional region $\mathcal S$ induced by the geometric description of the volume. Going through the motions of calculating this region; subsequently converting to cylindrical coordinates; and integrating with respect to $z$ and $r$ yields that the volume involves the single integral $$\int_0^{\pi/2} \frac{\cos \theta + \sin \theta - 1}{(\cos \theta + \sin \theta)^3} \, d \theta.$$ But after some thoughtful consideration, this integral proves quite difficult to evaluate by elementary methods: indeed, there is no obvious substitution because the derivative of $\cos \theta + \sin \theta$ is the difference (and not the sum) of the functions, and it seems that integration by parts is hopeless (for reasons I will expound upon momentarily). One immediate approach is to split the integral into two summands and try to evaluate each one in turn. $$\int_0^{\pi/2} \frac 1 {(\cos \theta + \sin \theta)^2} \, d \theta - \int_0^{\pi/2} \frac 1 {(\cos \theta + \sin \theta)^3} \, d \theta$$ We may evaluate the first integral by multiplying the numerator and denominator by $\sec^2 \theta$ and performing the substitution $u = 1 + \tan \theta.$ One can recognize the result as an improper integral that can be evaluated using the Fundamental Theorem of Calculus and L'Hôpital's Rule. But that leaves the second integral (and the object of my question here). $$\int_0^{\pi/2} \frac 1 {(\cos \theta + \sin \theta)^3} \, d \theta$$ Repeating the idea from above and multiplying and dividing by $\sec^3 \theta,$ we find that $$\int_0^{\pi/2} \frac 1 {(\cos \theta + \sin \theta)^3} \, d \theta = \int_0^{\pi/2} \frac{\sec^3 \theta}{(1 + \tan \theta)^3} \, d \theta.$$ Continuing via integration by parts with $u = \sec \theta,$ it follows that $$\int_0^{\pi/2} \frac{\sec^3 \theta}{(1 + \tan \theta)^3} \, d \theta = -{\left[\frac{\sec \theta}{2 (1 + \tan \theta)^2} \right]_0^{\pi/2}} + \frac 1 2 \int_0^{\pi/2} \frac{\sec \theta \tan \theta}{(1 + \tan \theta)^2} \, d \theta.$$ Once again, treating the first integral as improper, we may evaluate via the Fundamental Theorem of Calculus and L'Hôpital's Rule. Even still, it is the second integral that persists and continues to complicate matters.
Bearing all of this in mind, I am curious as to the "correct" approach to evaluate the integral $$\int_0^{\pi/2} \frac 1 {(\cos \theta + \sin \theta)^3} \, d \theta.$$ Based on a calculation from WolframAlpha, the inverse hyperbolic tangent function comes into play, but I am not sure how. I would appreciate any thoughtful commentary or consideration.
