Quintin Stafford

Answered

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.

$\int \mathrm{sin}(x)\mathrm{cos}(x)dx=\frac{1}{2}{\mathrm{sin}}^{2}(x)+{c}_{1}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}{\mathrm{cos}}^{2}(x)+{c}_{2}$

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)?

So here is the problem.

$\int \mathrm{sin}(x)\mathrm{cos}(x)dx=\frac{1}{2}{\mathrm{sin}}^{2}(x)+{c}_{1}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}{\mathrm{cos}}^{2}(x)+{c}_{2}$

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)?

Answer & Explanation

Donavan Scott

Expert

2022-07-02Added 22 answers

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

$(\frac{1}{2}{\mathrm{sin}}^{2}(x)+{c}_{1})-(-\frac{1}{2}{\mathrm{cos}}^{2}(x)+{c}_{2})$

which is equal to $\frac{1}{2}({\mathrm{sin}}^{2}(x)+{\mathrm{cos}}^{2}(x))+{c}_{1}-{c}_{2}$ and using the trig identity ${\mathrm{sin}}^{2}(x)+{\mathrm{cos}}^{2}(x)=1$, all of this is equal to just $\frac{1}{2}+{c}_{1}-{c}_{2}$ which is a constant.

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

$(\frac{1}{2}{\mathrm{sin}}^{2}(x)+{c}_{1})-(-\frac{1}{2}{\mathrm{cos}}^{2}(x)+{c}_{2})$

which is equal to $\frac{1}{2}({\mathrm{sin}}^{2}(x)+{\mathrm{cos}}^{2}(x))+{c}_{1}-{c}_{2}$ and using the trig identity ${\mathrm{sin}}^{2}(x)+{\mathrm{cos}}^{2}(x)=1$, all of this is equal to just $\frac{1}{2}+{c}_{1}-{c}_{2}$ which is a constant.

Oakey1w

Expert

2022-07-03Added 2 answers

Explanation:

Since $\text{}{\mathrm{sin}}^{2}(x)+{\mathrm{cos}}^{2}(x)=1$, if we let ${c}_{1}={c}_{2}-\frac{1}{2}$ so we'd have

$\begin{array}{rl}\frac{1}{2}{\mathrm{sin}}^{2}(x)+{c}_{1}& =\frac{1}{2}{\mathrm{sin}}^{2}(x)+({c}_{2}-\frac{1}{2})\\ & =\frac{1}{2}{\mathrm{sin}}^{2}(x)+({c}_{2}-\frac{1}{2})\\ & =\frac{1}{2}(1-{\mathrm{cos}}^{2}(x))+({c}_{2}-\frac{1}{2})\\ & ={c}_{2}-\frac{1}{2}{\mathrm{cos}}^{2}(x)\end{array}$

Since $\text{}{\mathrm{sin}}^{2}(x)+{\mathrm{cos}}^{2}(x)=1$, if we let ${c}_{1}={c}_{2}-\frac{1}{2}$ so we'd have

$\begin{array}{rl}\frac{1}{2}{\mathrm{sin}}^{2}(x)+{c}_{1}& =\frac{1}{2}{\mathrm{sin}}^{2}(x)+({c}_{2}-\frac{1}{2})\\ & =\frac{1}{2}{\mathrm{sin}}^{2}(x)+({c}_{2}-\frac{1}{2})\\ & =\frac{1}{2}(1-{\mathrm{cos}}^{2}(x))+({c}_{2}-\frac{1}{2})\\ & ={c}_{2}-\frac{1}{2}{\mathrm{cos}}^{2}(x)\end{array}$

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