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Homework Help
Calculus and Beyond Homework Help
Evaluate the integral (inverse trig functions)
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[QUOTE="ThatOneGuy, post: 4997596, member: 541432"] [h2]Homework Statement [/h2] [2[SUP]3/4[/SUP], 2] 4/(x√(x[SUP]4[/SUP]-4)) [h2]Homework Equations[/h2] ∫ du/(u√(u[SUP]2[/SUP] - a[SUP]2[/SUP])) = 1/a(sec[SUP]-1[/SUP](u/a) + c [h2]The Attempt at a Solution[/h2] I first multiplied the whole thing by x/x. This made the problem: 4x/(x[SUP]2[/SUP]√(x[SUP]4[/SUP] - 4)) Then I did a u substitution making u = x[SUP]2[/SUP]. Therefore, du = 2xdx. I multiplied by 2 to get 2du = 4xdx The problem then becomes 2∫du/(u√(u[SUP]2[/SUP] - 4)) Solving the integral I got 2[(1/2)sec[SUP]-1[/SUP](x[SUP]2[/SUP]/2)] from [2[SUP]3/4[/SUP], 2] I plug in the bounds and get 2[(1/2)sec[SUP]-1[/SUP](2) - (1/2)sec[SUP]-1[/SUP](2[SUP]6/4[/SUP]/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] [/QUOTE]
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Homework Help
Calculus and Beyond Homework Help
Evaluate the integral (inverse trig functions)
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