Trigonometric substitution, integral of 8dx/(4x^2+1)^2 basic error somewhere
Alright...been at this on and off all day!
The problem is the integral of 8dx/(4x^2+1)^2.
I let x=1/2*tanT
I work this problem, going through the integral of 4sec^2T/sec^4T
Push through to integral of 4*cos^2T, skipping forward I end up with 2t+sin(2t)
I build my triangle. Side opposite is 2x. Side adjacent is 1. Hypotenuse is 4x^2+1.
T=arctan(2x). Sin(2t) = side opposite (2x) over hypotenuse ( 4x^2+1 ).
So my answer is 2arctan(2x)+2x/(4x^2+1)+C However I have a problem. In the second term there, that should be 4x.
Any idea where I missed it?
Alright, second problem. Integrating x^2/(1-x^2)^(5/2)dx
Substitutions are x=sint
I work it all the way down to the integration of tan^2tsec^2t dt=tan^3t/3.
My triangle is side opposite=x, side adjacent= (1-x^2), hypotenuse 1. I substitute in and get (1/3)*(x/(1-x^2))^3 + C
Once again my side adjacent is messed up. The side adjacent in my answer should be the square root of what I have.
I think the problems are related.