How would you reconcile the following: ∫[(√u^2-a^2)/(u)] du = √(u^2-a^2)-acos^-1(a/|u|)+C AND ∫[(√u^2-a^2)/(u)] du = √(u^2-a^2)-asec^-1(|a|/u)+C . Use either integral to evaluate ∫√(e^(2x) − 6) dx.

Answer 1

Hi Kk T!

For the first part, draw a right triangle and label one of the acute angles A. Label the side adjacent to A with an a, and label the hypotenuse |u|.

cos(A) = a/|u| --> cos^(-1)(a/|u|) = A

sec(A) = |u|/a --> sec^(-1)(|u|/a) = A

So cos^(-1)(a/|u|) = sec^(-1)(|u|/a).

For the second part, let u = e^x. Then du = e^x dx, or dx = du/u. So your integrand is sqrt(u^2 - 6)/u. Now apply either of the given formulas, and then undo the substitution.

Hope that helps! Contact me if you would like to start tutoring. I am available.