∫(cos2x+1/sin4x )²

can any one help me with this step by step solution cause i have fill 2 papers of failed attempts(Headbang)

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- Dec 30th 2008, 08:51 PMmohamedsafyhelp bad integration problem
∫(cos2x+1/sin4x )²

can any one help me with this step by step solution cause i have fill 2 papers of failed attempts(Headbang) - Dec 30th 2008, 09:01 PMChop Suey
Is it $\displaystyle \int \left(\frac{\cos{2x}+1}{\sin{4x}}\right)^2~dx$

or

$\displaystyle \int \left(\cos{2x}+\frac{1}{\sin{4x}}\right)^2~dx$ - Dec 30th 2008, 09:57 PMmohamedsafy
it is the second one

∫(cos2x+(1/sin4x) )² - Dec 30th 2008, 10:04 PMChop Suey
Expand $\displaystyle \left(\cos{2x}+\frac{1}{\sin{4x}}\right)^2$ to get:

$\displaystyle \int \cos^2{2x}~dx + \int \frac{2\cos{2x}}{\sin{4x}}~dx + \int \csc^2{4x}~dx$

Use the identity $\displaystyle \cos^2{(2x)} = \frac{1+\cos{(4x)}}{2}$ and the first integral becomes trivial.

Use the identity $\displaystyle \sin{(4x)} = 2\sin{(2x)}\cos{(2x)}$ and the second integral becomes trivial.

The third is trivial as it is.