The definite integral of from 0 to 1 of ∫ (r3)dr/sqrt(4+r2)
∫udv = uv - ∫vdu
∫du/sqrt(a2 - u2) = arcsin(u/a) + C
∫du/(asqrt(a2 - u2)) = (1/a)arcsec(u/a) + C
The Attempt at a Solution
I made u = (4+r2)-1/2
because I thought it easier to get it's derivative, rather than integral by making it dv.
du = -r(4+r2)-3/2dr
dv = r3dr
v = (1/4)r4
I've set up the equation as
= uv -∫vdu
= r4/(4(4+r2)1/2) + ∫r3dr/(4+r2)3/2
This seems somewhat similar to the the second equation I put in "Relevant Equations" (where I could put a two in the numerator and multiply the integral by (1/2), but both top and bottom are to the power of 3 (which I don't know how to get rid of. AND the bottom is not a2 - u2 but a2 + u2.
If somebody could please help me understand this I would be greatly appreciative. I'd like to be able to master this type of integration, but I seem to have hit a wall.