# Thread: Show 2 functions have the same anti-derivative

1. 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).

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

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?

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

4. 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.