I've have no idea how to even approach this question. I've attempted to use $\frac{1}{5\tan^2(x)-1}$ but i'm really just grasping at straws here.
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In class you probally learned the concept of u substitution perhaps try the entire denominator – Faust Jan 14 '17 at 06:34
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Whats the closest question to this that you might know how to solve – jimjim Jan 14 '17 at 06:34
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You can use the method I posted here.http://math.stackexchange.com/questions/2095054/integrate-int-frac453cos2xdx/2095081#2095081 – Renascence_5. Jan 14 '17 at 06:37
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let $\displaystyle \mathcal{I} = \int \frac{1}{2-3 \cos 2x}dx = \int\frac{1}{2-3\left(\frac{1-\tan^2 x}{1+\tan^2 x}\right)}dx = \int \frac{\sec^2 x}{5\tan^2 x-1}dx$
substitute $\tan x= t$ and $\sec^2 xdx = dt$
so $\displaystyle \int\frac{1}{5t^2-1}dt = \frac{1}{5}\int \frac{1}{t^2-\frac{1}{\sqrt{5}}}dt = \frac{1}{2\sqrt{5}}\ln \bigg|\frac{\sqrt{5}t-1}{\sqrt{5}t+1}\bigg|+\mathcal{C}$
DXT
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This is good, for sure : you can still simplify the result and finally arrive to $\displaystyle \mathcal{I} =-\frac{\tanh ^{-1}\left(\sqrt{5} \tan \left(x\right)\right)}{\sqrt{5}}+C$ – Claude Leibovici Jan 14 '17 at 06:37
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3I don't understand what's occurring in the last two steps isn't $\displaystyle \int\frac{1}{5t^2-1}dt = \frac{1}{5}\int \frac{1}{t^2-\frac{1}{5}}dt$ and how did you get from that to the last step. – Julian Lopez Jan 14 '17 at 06:37
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1$ \frac{1}{5t^2-1} = \frac{1}{5(t^2-\frac{1}{5})} = \frac{1}{5} \frac{1}{t^2-\frac{1}{5}} $ – bigfocalchord Jan 14 '17 at 06:44