 Quintin Stafford

2022-07-01

How to reconcile the fact that the antiderivative of sin(x) cos(x) has two possible answers?
So here is the problem.
$\int \mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)dx=\frac{1}{2}{\mathrm{sin}}^{2}\left(x\right)+{c}_{1}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}{\mathrm{cos}}^{2}\left(x\right)+{c}_{2}$
This fact doesn't make much sense to me. How do you reconcile the fact that there are two possibilities (notice that these answers aren't only different by a constant)?
What are some other functions that have more than one antiderivatives (not only different by a constant)? Donavan Scott

Expert

Explanation:
You wrote:
(notice that these answers aren't only different by a constant)
But they do differ by a constant. If you subtract one solution from the other, you have
$\left(\frac{1}{2}{\mathrm{sin}}^{2}\left(x\right)+{c}_{1}\right)-\left(-\frac{1}{2}{\mathrm{cos}}^{2}\left(x\right)+{c}_{2}\right)$
which is equal to $\frac{1}{2}\left({\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)\right)+{c}_{1}-{c}_{2}$ and using the trig identity ${\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)=1$, all of this is equal to just $\frac{1}{2}+{c}_{1}-{c}_{2}$ which is a constant. Oakey1w

Expert

Explanation:
Since , if we let ${c}_{1}={c}_{2}-\frac{1}{2}$ so we'd have
$\begin{array}{rl}\frac{1}{2}{\mathrm{sin}}^{2}\left(x\right)+{c}_{1}& =\frac{1}{2}{\mathrm{sin}}^{2}\left(x\right)+\left({c}_{2}-\frac{1}{2}\right)\\ & =\frac{1}{2}{\mathrm{sin}}^{2}\left(x\right)+\left({c}_{2}-\frac{1}{2}\right)\\ & =\frac{1}{2}\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)+\left({c}_{2}-\frac{1}{2}\right)\\ & ={c}_{2}-\frac{1}{2}{\mathrm{cos}}^{2}\left(x\right)\end{array}$

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