Show 2 functions have the same anti-derivative

[NoWay1]

So I have to show 2sin^2(x) and -cos(2x) have the same antiderative.

Here's how I approached this.

2sin^2(x) = 1-cos2x ==> u = 2x
intergral of that is
(u - sinu)/2 + c = x - (sinx)/2 + c

-cos2x ==> u = 2x
intregal of that is
(-sinu)/2 + c= -(sin2x)/2 + c

Have I calculated/approached this exercise wrongly? I don't see how they could be the same antiderative
I have another similar exercise with 2cos^2(x) and cos(2x).

 

[MarkFL]

Since this question involves integral calculus, I have moved it here. :D

In order for two functions to have the same anti-derivative, they must in fact be equivalent. From a double-angle identity for cosine, we know:

\(\displaystyle \cos(2x)=1-2\sin^2(x)\)

Hence:

\(\displaystyle -\cos(2x)=2\sin^2(x)-1\ne2\sin^2(x)\)

Thus, the two given functions are not equivalent, therefore they cannot have the same anti-derivative.

Can you use a similar line of reasoning for the second question?

 

[NoWay1]

Thanks a lot for the confirmation, way the question was presented was very confusing to me, as if they had to have the same antideravative

 

[Prove It]

I think the question actually must be to show that the two functions have the same derivative.

As $\displaystyle \begin{align*} -\cos{(2\,x)} \equiv -\left[ 1 - 2\sin^2{(x)} \right] \equiv 2\sin^2{(x)} -1 \end{align*}$, which only differs from $\displaystyle \begin{align*} 2\sin^2{(x)} \end{align*}$ by a constant, they will in fact have the same derivatives.

 

[NoWay1]

This must be the answer I needed, thank you