Getting two different integrals for same function(?)

[1/2"]

Homework Statement

The actual problem is ∫sin2x/((sinx)⁴+(cosx)⁴) dx

Homework Equations

The Attempt at a Solution

First wrote the expression as
∫$\frac{2sin2x}{((sinx)^2+(cosx)^2)^2+((sinx)^2-(cosx)^2))^2 }$ dx
then I changed the 2dx to d(2x)
∫$\frac{sin2x}{((sinx)^2+(cosx)^2)^2+((sinx)^2-(cosx)^2))^2 }$ d2x

∫$\frac{sin2x}{1+ (cos2x)^2 }$ d2x

Multiplying the whole expressioin with (sec2x)^2

∫$\frac{tan2x sec2x}{1+ (sec2x)^2 }$ d2x

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

taking sec2x =t

∫$\frac{1}{1+ t^2 }$ 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 :
- ∫-$\frac{1}{1+ t^2 }$ dt

- arccot t

-arctan (cos2x)

Am i going wrong somewhere? please could you point it out ?
Thanks in advance!

 

[haruspex]

When an indefinite integral appears to give two different answers, it generally turns out that the difference is a constant. I believe that is the case here.

 

[1/2"]

Actually it has got limits - 0 and $\frac{∏}{2}$

And putting the limits the answer in the first case come as $\frac{∏}{2}$ and in the second one -$\frac{∏}{2}$

When an indefinite integral appears to give two different answers, it generally turns out that the difference is a constant. I believe that is the case here.

Ya the difference is a constant arctan ∞ = $\frac{∏}{2}$

If I happen to give either of the answer will it be marked correct ? or do I have to work out both of them ?

 

[haruspex]

Actually it has got limits - 0 and $\frac{∏}{2}$

And putting the limits the answer in the first case come as $\frac{∏}{2}$ and in the second one -$\frac{∏}{2}$

With those limits, there will be some doubt about the validity since your integral goes through an infinity at π/4. You could fix that by observing there's symmetry about π/4, so you can take the integral 0 to π/4 and double it.
However, I don't see why you are getting different answers. When I substitute those limits in your integrals I get $\frac{∏}{2}$ for both.