Is there any nonlinear function $f(x)$ for which both $\sin (f(x))$ and $\cos(f(x))$ have closed-form integrals?

Question

I'm interested in closed-form antiderivaties of elementary functions.

The question arises from seeing that something as simple as $\int\sin(x^2)dx$ and $\int\cos(x^2)dx$ generates the Fresnel integrals.

I'm familiar with integration of elementary functions that yields elementary functions again.

Answer 1

Yes.
There are a lot of functions that satisfy this condition.

We have:

$$\sin(f(x))=-\frac{1}{2}ie^{if(x)}+\frac{1}{2}ie^{-if(x)},$$ $$\cos(f(x))=\frac{1}{2}e^{if(x)}+\frac{1}{2}e^{-if(x)}.$$

We see, the right-hand sides of these formulas differ from each other only in the constants. The antiderivative of a product of a constant factor and a non-constant term is the product of the constant factor and the antiderivative of the non-constant term. If $\sin(f(x))$ has an antiderivative in closed form, $\cos(f(x))$ also has an antiderivative in closed form therefore which differs only in the constant factors - and vice versa. That means $\sin(f(x))$ has an antiderivative in closed form iff $\cos(f(x))$ also has an antiderivative in closed form.

The antiderivative of a sum is the sum of the antiderivatives of its summands. We therefore only need to determine the antiderivatives of $e^{if(x)}$ and $e^{-if(x)}$.

Also for elementary $f(x)$, the antiderivative of $e^{f(x)}$ is often not an elementary function. See e.g. in the references Smith and Conjectures 4 and 5 of Yadav.
Let's use therefore a simple way to find some elementary functions $f(x)$: we set $f(x)=-i\ln(g(x))$ and get $e^{\pm if(x)}=\pm g(x)$.

Because the antiderivatives of $e^{f(x)}$ often are nonelementary integrals, certain Special functions were introduced that are nonelementary antiderivatives of elementary functions. Examples are the following Liouvillian functions: elliptic integrals, logarithmic integrals, error function, Gaussian integrals, Fresnel integrals.

For deciding which kinds of functions can have a closed-form antiderivative, we also have the specialist area Symbolic integration with Liouville's theorem and Risch algorithm for elementary functions and Liouvillian functions.

You can play with functions at Wolfram Alpha.
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[Smith] TR Smith: List of Functions Without Antiderivatives
[Yadav 2012] Yadav, D. K.: A Study of Indefinite Nonintegrable Functions. PhD thesis, Vinoba Bhave University, India, 2012
[Yadav 2016-1] Yadav, D. K.: Six Conjectures in Integral Calculus. 2016
[Yadav 2016-2] Yadav, D. K.: Six Conjectures on Indefinite Nonintegrable Functions or Nonelementary Functions. 2016

Answer 2

Here is a perhaps not-so-obvious example: let $f(x) = \ln(x).$ Then

\begin{align} \int \sin(\ln(x))\,\mathrm dx &= -\frac12 x (\cos(\ln(x)) - \sin(\ln(x))) + C, \\ \int \cos(log(x))\,\mathrm dx &= \frac12 x (\sin(\ln(x)) + \cos(\ln(x))) + C. \end{align}

Answer 3

As mentioned in the comments taking $f$ equal to the inverse sine or cosine function works wonders. Here is another example.

Take $f(x)=x^2+1$ then

$$\int\sin(x^2+1)dx=\int \cos(1)\sin(x^2)+\sin(1)\cos(x^2)dx$$

Substituting in $u=\frac{\sqrt{2}x}{\sqrt{π}}$ makes it easy to see that

$$\int\sin(x^2+1)dx=\frac{\sqrt{π}(\cos(1)S(\frac{\sqrt{2}x}{\sqrt{π} })+\sin(1)C(\frac{\sqrt{2}x}{\sqrt{π}}))}{\sqrt{2}}+C$$

Similarly we get that

$$\int\cos(x^2+1)dx=-\dfrac{\sqrt{{\pi}}\left(\sin\left(1\right)\operatorname{S}\left(\frac{\sqrt{2}\,x}{\sqrt{{\pi}}}\right)-\cos\left(1\right)\operatorname{C}\left(\frac{\sqrt{2}\,x}{\sqrt{{\pi}}}\right)\right)}{\sqrt{2}}$$

Where $C,S$ are the Fresnel integrals

Also notice that $\frac{d}{dx}(x^2+1)=2x$ , which is $0$ when $x$ is $0$ . So the answer to the question " Is there any function $f(x)$ whose derivative can be $0$ somewhere and still gives a closed form integral?" Is yes .