How do you integrate 1e2x2ex24dx using trigonometric substitution?

1 Answer
Jun 11, 2018

Use the substitution 24ex+1=5sinθ.

Explanation:

Let

I=1e2x2ex24dx

Rewrite in terms of ex:

I=ex12ex24e2xdx

Complete the square in the square root:

I=24ex25(24ex+1)2dx

Apply the substitution 24ex+1=5sinθ:

I=2415cosθ(524cosθdθ)

Simplify:

I=124dθ

The integral is trivial:

I=124θ+C

Reverse the substitution:

I=124sin1(24ex+15)+C