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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!