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I am looking for a solution for the following integral problem.

$$\int \sin(x)\cos(3x^2)dx$$

Passed over these integral things long time ago. I cannot see how to go for a solution.

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    $\begingroup$ There is no simple expression for this integral in terms of elementary functions. $\endgroup$
    – Simon S
    Apr 21, 2015 at 15:25
  • $\begingroup$ I took off the erroneously tagged tag "indefinite integral"; for what you are looking for is a general form of the primitives. $\endgroup$
    – Yes
    Apr 21, 2015 at 15:38
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    $\begingroup$ Look for "Fresnel Integrals" and "Error Function" $\endgroup$
    – tired
    Apr 21, 2015 at 15:52
  • $\begingroup$ Is Wolfram Alpha down? I checked: it's not $\endgroup$
    – user147263
    Apr 21, 2015 at 18:22

2 Answers 2

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With some work can express the integral as series in terms of $\sin(3x^2)$ and $\cos(3x^2)$. However, the terms will take a while to work out. This will give you some approximate solutions to the integral. Luckily, the oddness of $\sin(x)$ works nicely with $\cos(3x^2)$ for integration.

We can start with the series representation of $$\sin(x)=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$ and plug this into your integral equation.

This leads to $$\int \left(\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}\right) \cos(3x^2)dx = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\int x^{2n+1} \cos(3x^2) dx. $$

From here, we need to integrate each one of the integrals $$P_{2n+1}(x) = \int x^{2n+1} \cos(3x^2) dx.$$ Therefore the integral is given by $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!} P_{2n+1}(x).$$

Let's call $$Q_{2n+1}(x) = \int x^{2n+1} \sin(3x^2) dx.$$

The first few are easy to integrate: $$P_1(x) = \int x \cos(3x^2) dx = \frac16 \sin(3x^2)\text{, and } Q_1(x) = \int x \sin(3x^2)dx = -\frac16 \cos(3x^2).$$

Then $$P_3(x) = \int x^3 \cos(3x^2) dx$$ can be broken up by integration by parts, taking $u = x^2$ and $dv = x \cos(3x^2) dx$. Thus

$$P_3(x) = x^2 P_1(x) - \int 2xP_1(x)dx = x^2 P_1(x) - \frac13 \int x\sin(3x^2)dx = x^2 P_1(x) - \frac13 Q_1(x).$$

In other words $$P_3(x) = \frac{x^2\sin(3x^2)}{6} + \frac{\cos(3x^2)}{18}.$$

We can calculate $P_5(x)$ again using integration by parts, indeed $$P_5(x) =\int x^{5} \cos(3x^2) dx = x^4 P_1(x) - \int P_1(x) 4x^3 dx$$ $$=x^4 P_1(x) - \frac46 \int x^3 \sin(3x^2) dx$$

$$=x^4 P_1(x) - \frac46 Q_3(x).$$

Note that we have to also calculate $Q_3(x)$ to find $P_5(x)$. This can be done the same way that $P_3(x)$ was integrated. Using the first two terms we calculated, this gives:

$$\int \sin(x)\cos(3x^2)dx \approx \frac16 \sin(3x^2) - \frac{1}{6} \left( \frac{x^2\sin(3x^2)}{6} + \frac{\cos(3x^2)}{18} \right),$$ which will be a good approximation when $x$ is near zero.


We can work out a general recursion for $Q_{2n+1}(x)$ and $P_{2n+1}(x)$ by writting $$P_{2n+1}(x) = \int x^{2n+1} \cos(3x^2) dx = \frac{1}{6} x^{2n} \sin(3x^2) - \frac{n}{3} \int x^{2n-1} \sin(3x^2)dx.$$ So $$P_{2n+1}(x)= \frac16 x^{2n} \sin(3x^2) - \frac{n}{3} Q_{2n-1}(x),$$ and

$$Q_{2n+1} = -\frac{1}{6} x^{2n} \cos(3x^2) + \frac{n}{3} P_{2n-1}(x).$$

Plugging in $Q_{2n-1}$ into $P_{2n+1}$ we find:

$$P_{2n+1} = \frac16 x^{2n} \sin(3x^2) - \frac{n}{3} \left( -\frac{1}{6} x^{2n-2} \cos(3x^2) + \frac{n-1}{3} P_{2n-3}(x) \right)$$

$$=\frac16 x^{2n} \sin(3x^2) + \frac{n}{18} x^{2n-2} \cos(3x^2) - \frac{n(n-1)}{9} P_{2n-3}(x).$$

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$$\begin{align}I=\dfrac12~\sqrt{\dfrac\pi6}~\bigg\{\cos\dfrac1{12}\bigg[S\bigg(\dfrac{6x+1}{\sqrt{6\pi}}\bigg)-S\bigg(\dfrac{6x-1}{\sqrt{6\pi}}\bigg)\bigg]+\\\\+\sin\dfrac1{12}\bigg[C\bigg(\dfrac{6x-1}{\sqrt{6\pi}}\bigg)-C\bigg(\dfrac{6x+1}{\sqrt{6\pi}}\bigg)\bigg]\bigg\}\end{align}$$

where S and C are the two Fresnel integrals.

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