Here's what I have done:
Visualizing a right triangle with the horizontal side = 1 and the vertical side = x,
I thought that I had this, but the answer doesn't check out. Can anybody see what I am missing? Thanks again!
When you make the substitution , you are converting the integrand to a function of , which means your terminals need to be values of , not .
So it should be .
But on an entirely different note, I believe hyperbolic substitution will be easier than trigonometric in this case.
Let .