How do you integrate ∫1√e2x−2ex−24dx using trigonometric substitution?
1 Answer
Jun 11, 2018
Use the substitution
Explanation:
Let
I=∫1√e2x−2ex−24dx
Rewrite in terms of
I=∫e−x√1−2e−x−24e−2xdx
Complete the square in the square root:
I=√24∫e−x√25−(24e−x+1)2dx
Apply the substitution
I=√24∫15cosθ(−524cosθdθ)
Simplify:
I=−1√24∫dθ
The integral is trivial:
I=−1√24θ+C
Reverse the substitution:
I=−1√24sin−1(24e−x+15)+C