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I am not even going to show my work because I have accomplished nothing attempting this problem. I have absolutely no idea what to do.

$$\int \frac{\cos^5 x}{ \sqrt {\sin x}}\, dx$$

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    It's an application of the rule of thumb needed for this and several of your previous questions: If you have $\cos$ to an odd power upstairs in the integrand, try factoring out one $\cos$ term, write the other $\cos$ terms in terms of $\sin$, and let $u=\sin x$.2012-06-03
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    @DavidMitra I have memorized that rule because it is in a table in my book. I know how to do simple problems like that but that was not my problem.2012-06-03
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    This silly system should tell me that someone else is editing the post at the same time.2012-06-03

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Write $\cos^{5}(x) = (1-\sin^{2}{x})^{2} \cdot \cos{x}$ and then put $t =\sin{x}$.

So you will have \begin{align*} \int \frac{\cos^{5}(x)}{\sqrt{\sin{x}}} \ dx &= \int \frac{(1-\sin^{2}(x))^{2}}{\sqrt{\sin{x}}} \cdot \cos{x} \ dx \\\ &=\int \frac{(1-t^{2})^{2}}{t^{1/2}} \ dt \\\ &=\int \frac{1-2t^{2}+t^{4}}{t^{1/2}} \ dt \end{align*}

Don't forget to then substitute u= t^(1/2) so du = (1/2)u^(-1/2)

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    How do I know that is the correct way of doing it? Out of the millions of ways to do this problem, why chose this way? I did it about 12 other ways and I never considered this way of doing it. Also I think your math is wrong, it seems to be missing the dt subsitution, shouldn't there be a 1/2?2012-06-03
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    @jordan.: See when you have $\sin{x}$ in the denominator and cos(x) in the numerator then intuition tells that i have to put $t=\sin(x)$ to get rid of the denominator. Now all that is left is how i can get a $\cos(x)$ term up2012-06-03
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    But how does that get rid of the denominator? Or is this a rule I just need to memorize like the others? I was trying to transform the problem without using that.2012-06-03
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    Also this seems wrong, how can I work with a cosx and a sint. Is that legal to work with two variables?2012-06-03
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    @jordan: Getting rid of the denominator means making denominator as $t$ so that you can work with. You don't need to memorize. Just practise, think for sometime and then see if you don't get.2012-06-03
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    @jordan: Please understanD. If sin(x) is t then dt is cos(x) and everything changes. there is nothing wrong here.2012-06-03
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    That is wrong, there should be a 1/cosx somewhere.2012-06-03
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    @Jordan why should there be a $1/cos(x)$2012-06-03
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    Because if t = sinx then dt/cosx = dx2012-06-03
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    If $t = \sin x$ then $$\frac{\mathrm{d}t}{\mathrm{d}x} = \cos x \ \Rightarrow \ \mathrm{d}t = \cos x \mathrm{d}x$$.2012-06-03
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    no: dt = \cos(x) dx2012-06-03
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    That is the same thing isn't it?2012-06-03
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    @Jordan you seem to have a rather negative attitude towards mathematics, and it is clear you don't like it and that you do it only because you have to. I'm sorry this is your situation as many of us get lost of fun, pleasure and intellectual challenge from mathematics, but if you ***have to do it*** then you might as well try a little harder and make a supreme effort to begin having a more positive attitude towards this stuff. That way you'll enjoy more (or will suffer less) and I'm sure you'll begin having more success.2012-06-03
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    @DonAntonio I honestly do not have the time to enjoy this, it is something I have to do and it is something that I do not have time to do. I just have to put my hours in and try and do it.2012-06-03
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    @Jordan Fair enough, but then don't get pesky at people doing their best to help you. And try to find some good time to invest in this, as it is my experience that people with an attitude simmilar to yours are in great danger of failing exams and stuff.2012-06-03
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    @DonAntonio I put in about 10 hours a day.2012-06-03
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    @Chandrasekhar: there is a missing power in the second integral, i think.2012-06-03
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    If that's true then you're doing it wrong. If you're looking for things to memorize then it would probably take that long, if you're looking for patterns and an understanding of what exactly these transformations achieve then it won't. Every time you solve a problem, or see an answer, look at what makes that answer work and try to come up with a new problem that can be solved that way. You need to see underneath the symbol pushing if you want any kind of payoff. Studying smarter is more important than studying harder.2012-06-03
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    @chris: Thanks. Yeah2012-06-03
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    @Jordan I have been through this myself and have seen hundreds of my own students go through this. Yes math is hard. You don't like it fine. You don't have to like it but you need it to get to wherever it is you want to go so just be good at it. And if it is any consolation, math is difficult for people who do love it. And in order to be good at it, memorization is not going to turn you into a good mathematician. Memorization has its uses but understanding and comprehension is much more important.2012-12-07
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    @Jordan Not trying to put you down or anything but if memorization of tables is your strategy, I guarantee you before you reach the end of calculus you will be completely overwhelmed and you brain will be refused to crammed with useless boring formulas which it doesn't understand and doesn't know how/when to use. It takes patience, practice, and perseverance. That's it. No shortcuts!2012-12-07