You need 3x = 2tanθ, so
9x^2 = 4tan^2 θ
9x^2+4 = 4tan^2 θ + 4 = 4sec^2 θ
Hello, I have a question concerning trigonometric substitution.
let's say we have
integral of dx/sqrt(9x^2 + 4),
so after doing a few steps we have:
2/3 integral of sec0/sqrt((2tan0)^2 + 4*) d0 (the * is for later on)
the next step turns into:
2/3 integral of sec0/(sqrt(4tan^2 0 + 1))
I know that the sqrt(4 tan^2 0 + 1) is because of the power of 2 from the first step, but where does the other + 4 disappear (the one I put a * next to).
2 answers
thank you Steve