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Thread: indefinite trigonometric integral

  1. #1
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    indefinite trigonometric integral

    $\displaystyle \int \frac {\cos^5 x}{sqrt(sin x)} dx$

    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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  2. #2
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    Re: indefinite trigonometric integral

    Look at $\displaystyle \int \frac{(\cos^2(x))^2(\cos(x))}{\sqrt{\sin(x)}}dx$.

    I get there by saying to myself: "I'd *like* to substitute $\displaystyle u = \sin(x)$ to simply that ugly denominator. That means I need $\displaystyle du = \cos(x)dx$. OK, I'll peel one of those cosines in the numerator off and put it with the $\displaystyle dx$. 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."
    Last edited by johnsomeone; Oct 1st 2012 at 11:29 AM.
    Thanks from MarkFL
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