Solve the following problem using trig substitution:
Hello, drewbear!
This problem is quite awful!
Under the radical, complete the square:
. .
The integral becomes: .
Let
Substitute: .
. . . . .
. . . . .
. . . . .
Back-substitute: .
. .
. .
We have: .
. . . . . .
This can be simplified further, but I won't bother . . .