What must be true of F(x) and G(x) if both are antiderivatives of f(x)?

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sonorous9n · Accepted Answer

Step 1  There must exists a constant C such that   F(x)−G(x)=C  In other words two antiderivatives of a function can differ only by a constant.

veiga34 · Accepted Answer

Step 1
If F(x) and G(x) both are antiderivatives of f(x).
Then F(x) can be made to equal G(x), by adding or subtracting some constant. i.e. They differ by a constant
For example,
Take $\int \sin x \cos x d x$
Way 1:
Put $\sin x = u$
So, that $\cos x d x = d u$
Therefore,
$\int \sin x \cos x d x = \int u d u$
$= \frac{u^{2}}{2} + C$
Reverting back the substitution,
$\int \sin x \cos x d x = \frac{\sin^{2} x}{2}$
$\int \frac{\sin 2 x}{2} d x$
Put x=t, so that 2dx=dt
$\int \frac{\sin 2 x}{2} d x = \frac{1}{4} \int \sin t d t$
$= \frac{- \cos t}{4} + C$
Reverting back the substitution, we get
$\int \frac{\sin 2 x}{2} d x = \frac{- \cos 2 x}{4} + C_{2}$
Now , we got two seemingly different anti-derivatives for our integral
But , now using some trigonometric identities , we can prove that they differ only by a constant.
We know $\cos 2 x = 1 - 2 \sin^{2} x$
Therefore, $\frac{- \cos 2 x}{4} + C_{2} = \frac{2 \sin^{2} x - 1}{4} + C_{2}$
$= \frac{\sin^{2} x}{2} - \frac{1}{4} + C_{2}$
Since, $C_{2}$ is constant, $- \frac{1}{4}$ is constant, $C_{1}$ is also constant. Thus, they differ only by a constant.

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