Function different from original by differentiating it and integrating it again.

Question

Let a function be defined $f(x)=sin^4x+cos^4x,$ $x\in R$.
Rewriting this, $$f(x)=(sin^2x+cos^2x)^2-2sin^2xcos^2x$$$$\implies f(x)=1-\frac{sin^22x}{2}$$$$\implies \frac{1}{2}\leq f(x)\leq1 $$

However,upon differentiating the function,we get$$f'(x)=4sin^3x cosx-4cos^3xsinx$$$$\implies f'(x)=4cosxsinx(sin^2x-cos^2x)$$$$\implies f'(x)=-2sin2xcos2x$$$$\implies f'(x)=-sin4x$$Integrating both sides,$$f(x)=\frac{cos4x}{4}$$which lies between$\frac{-1}{4}$ and $\frac{1}{4}$and is not equal to the original function.Why is the answer different?

Answer 1

Yes, $f'(x)=-\sin(4x)$. But $f(0)=1$ and the primitive of $-\sin(4x)$ which maps $0$ into $1$ is $\frac34+\frac14\cos(4x)$, whose range is $\left[\frac12,1\right]$.

Answer 2

I graphed $x\mapsto \sin^4 x + \cos^4 x$ and $x \mapsto \frac{1}{4}\cos4x$ on the same axes, and here's what I saw.

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Answer 3

You have to remember that for a constant $k$, we have that $\frac{d}{dx} (k) =0$. Hence when we integrate a function in different ways, we may obtain functions which differ by a constant. Differentiating once more will see you obtain the same function.