help bad integration problem

**mohamedsafy** · Dec 30th 2008, 08:51 PM

∫(cos2x+1/sin4x )²

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

**Chop Suey** · Dec 30th 2008, 09:01 PM

Is it $\int \left(\frac{\cos{2x}+1}{\sin{4x}}\right)^2~dx$

or

$\int \left(\cos{2x}+\frac{1}{\sin{4x}}\right)^2~dx$

**mohamedsafy** · Dec 30th 2008, 09:57 PM

it is the second one

∫(cos2x+(1/sin4x) )²

**Chop Suey** · Dec 30th 2008, 10:04 PM

Expand $\left(\cos{2x}+\frac{1}{\sin{4x}}\right)^2$ to get:

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

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

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

The third is trivial as it is.

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