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## Homework Statement

The actual problem is ∫sin2x/((sinx)

^{4}+(cosx)

^{4}) dx

## Homework Equations

## The Attempt at a Solution

First wrote the expression as

∫[itex]\frac{2sin2x}{((sinx)^2+(cosx)^2)^2+((sinx)^2-(cosx)^2))^2 }[/itex] dx

then I changed the 2dx to d(2x)

∫[itex]\frac{sin2x}{((sinx)^2+(cosx)^2)^2+((sinx)^2-(cosx)^2))^2 }[/itex] d2x

∫[itex]\frac{sin2x}{1+ (cos2x)^2 }[/itex] d2x

Multiplying the whole expressioin with (sec2x)^2

∫[itex]\frac{tan2x sec2x}{1+ (sec2x)^2 }[/itex] d2x

tan2xsec2x d(2x) = d((sec2x)^2)

taking sec2x =t

∫[itex]\frac{1}{1+ t^2 }[/itex] dt

__At this stage I am getting 2 different result when i proceed in 2 different way__

__First one:__

integrating i get

arctan t = arctan (sec 2x)

__Second one :__

- ∫-[itex]\frac{1}{1+ t^2 }[/itex] dt

- arccot t

-arctan (cos2x)

Am i going wrong somewhere? please could you point it out ?

Thanks in advance!