# Evaluating an Trigonometric Integral

I'm completely at a loss on how to solve this integral. I've tried to put it in programs like Wolfram Alpha and Microsoft Maths to get some semblance of an answer to aim for but all I get is a garbled mess.

To my knowledge it doesn't fit into any trigonometric identities that I know, and no other methods such as integration by parts and substitution don't really fit. So I'm just looking for some pointers on where to start.

$$\int \! \frac {\cos(3x)}{\sqrt{1-\sin(3x)}} dx$$

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## 3 Answers

$\bf Hint:$ make the sustitution $u=1-\sin(3x)$.

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Or even $u=\sqrt{1-\sin3x}$. – Gerry Myerson Feb 26 '12 at 23:03
Ah that makes sense thanks, I was only looking at substituting in terms of x and not the whole root – Amy Feb 26 '12 at 23:44
I got to $$\int \! \frac {\cos(3x)}{\sqrt{u}} * -\frac {1}{3cos(3x)} du$$ but know how to proceed :s – Amy Feb 27 '12 at 0:14
@amy Cancel the $\cos$ terms; then write ${1\over\sqrt u}=u^{-1/2}$ – David Mitra Feb 27 '12 at 1:01

Here's how I sometimes state a hint for problems of this kind: $$\int \frac{\cos(3x)}{\sqrt{1-\sin(3x)}}\;dx = \int \frac{1}{\sqrt{1-\sin(3x)}}{\huge(} \cos(3x)\;dx{\huge)}.$$ That's the hint.

The point is that the expression inside the huge parentheses should become $du$ or a constant times $du$. And that's just what you should be looking for when you do problems of this kind.

So letting $u=1-\sin(3x)$ we get $du=-3\cos(3x)$, so the integral is $$\int \frac{1}{\sqrt{u}}\; \left(\frac{-1}{3}\;du\right) = \frac{-1}{3}\int \frac{1}{\sqrt{u}}\;du.$$

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$$\frac{\cos{3x}}{\sqrt{1-\sin{3x}}}=\frac{\cos{3x}\sqrt{1+\sin{3x}}}{\sqrt{1-\sin{3x}}\sqrt{1+\sin{3x}}}=\sqrt{1+\sin{3x}}$$

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How does this help with evaluating the integral? – Hans Lundmark Feb 27 '12 at 7:20