I haven't been able to figure out a way to deal with the square root in the denominator.

The solution the book provides makes it clear that i should be using a Pythagorean identity in the usual way.

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- Oct 1st 2012, 10:36 AMbkbowserindefinite trigonometric integral

I haven't been able to figure out a way to deal with the square root in the denominator.

The solution the book provides makes it clear that i should be using a Pythagorean identity in the usual way. - Oct 1st 2012, 11:22 AMjohnsomeoneRe: indefinite trigonometric integral
Look at .

I get there by saying to myself: "I'd *like* to substitute to simply that ugly denominator. That means I need . OK, I'll peel one of those cosines in the numerator off and put it with the . If I'm lucky, the rest of the numerator can be expressed in terms of u. Looking at it, I see that I am indeed lucky."