Multiple answer for integration of a function?

Question

Q. $\int \left(\frac{sin2x}{sin^4x+cos^4x}\right)\:dx$

My method:

$$\int \:\left(\frac{sin2x}{sin^4x+cos^4x}\right)\:dx=\int \:\:\left(\frac{sin2x}{\left(cos^2x-sin^2x\right)^2+2sin^2\left(x\right)cos^2\left(x\right)}\right)\:dx\:$$

=> $$\int \:\left(\frac{2sin2x}{2\left(cos^2\left(2x\right)\right)+sin^2\left(2x\right)}\right)=\int \:\left(\frac{2sin2x}{cos^2\left(2x\right)+1}\right)$$

Now let $t=cos\left(2x\right)$

therefore $-dt=2sin\left(2x\right)$

So we have the integration as:

$$\int \:\left(\frac{-dt}{t^2+1}\right)=-\tan ^{-1}\left(t\right)=-\tan \:^{-1}\left(cos\left(2x\right)\right)$$

But my text provides the answer as: $\tan \:^{-1}\left(\tan ^2\left(x\right)\right)$. Am I wrong somewhere?

Answer 1

$$\frac{\sin2x}{\sin^4x+\cos^4x}=2\frac{\tan x\sec^2x}{1+\tan^4x}$$

Set $u=\tan^2x$ to find $I=\arctan(\tan^2x)=\arctan\dfrac{1-\cos2x}{1+\cos2x}=\arctan1-\arctan(\cos2x)$

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$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^22x}2=1-\frac{1-\cos^22x}2$$

Set $\cos2x=u$

Answer 2

Hint

What you did is very good.

The two expressions just differ by $\frac{\pi}{4}$ so you call select the integration constant (you forgot it) to go from one to the other.