Well, you can skip steps if you like. That's going too fast for the likes of me. I would just substitute \(\displaystyle v=n+0.5\) first, since you don't gain any complexity.

I do trig substitutions all the same way. First step: draw a triangle. Next step, assign names to sides. If the squares in the integrand are added, make the hypotenuse equal to the sum of the squares of the other two sides. If the squares have opposite sign, then make the positive one the hypotenuse, and one of the other sides gets to be the negative one. Assign an angle somewhere, and you're off and running. The virtue of doing it this way, is that often the inverse substitution (for indefinite integrals, this is required) is quite apparent straight from the triangle.