Evaluate the integral (inverse trig functions)

Homework Statement

[23/4, 2] 4/(x√(x4-4))

Homework Equations

∫ du/(u√(u2 - a2)) = 1/a(sec-1(u/a) + c

The Attempt at a Solution

I first multiplied the whole thing by x/x. This made the problem:
4x/(x2√(x4 - 4))

Then I did a u substitution making u = x2. Therefore, du = 2xdx. I multiplied by 2 to get 2du = 4xdx

The problem then becomes 2∫du/(u√(u2 - 4))

Solving the integral I got 2[(1/2)sec-1(x2/2)] from [23/4, 2]

I plug in the bounds and get 2[(1/2)sec-1(2) - (1/2)sec-1(26/4/2)

This is where I'm lost. The second sec does not seem like a nice number and I'm assuming my professor would make the problem come out nicely as he always has. I'm pretty sure I made a mistake somewhere because of this but I don't know where.[/B]

jambaugh
Gold Member
Try simplifying the fractional exponent and rewriting inverse secant in terms of inverse cosine ( arcsec(x) = arccos(1/x)).

ThatOneGuy
Try simplifying the fractional exponent and rewriting inverse secant in terms of inverse cosine ( arcsec(x) = arccos(1/x)).

Thank you so much! I should have noticed the reducing.

Finishing the problem:

sec-1(2) - sec-1(21/2)

sec-1(2) - sec-1(√(2))

This equals:
arccos(1/2) - arccos(1/√(2))
arccos(1/2) - arccos(√(2)/2)
pi/3 - pi/4 = pi/12