Evaluate $\int_0^\pi \sin^4\left(x+\sin 3x\right)dx$

My working let I=$\int_0^\pi \sin^4\left(x+\sin 3x\right)dx$

$=\int_0^\pi \frac18\left(\cos (4x+4\sin 3x)-4\cos(2x+2\sin 3x)+3\right)dx$

$=\frac{3\pi}{8}+\frac18\int_0^\pi\cos (4x+4\sin 3x)dx-\frac12\int_0^\pi\cos (2x+2\sin 3x)dx$

  • 1
    $\begingroup$ What leads you to believe that there is a "nice" closed form answer? $\endgroup$ – Mark Viola Jul 5 '18 at 18:17
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    $\begingroup$ I think you will need a numerical method. $\endgroup$ – Dr. Sonnhard Graubner Jul 5 '18 at 18:25
  • $\begingroup$ it is $\approx 1.1781$ $\endgroup$ – Dr. Sonnhard Graubner Jul 5 '18 at 18:26
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    $\begingroup$ I believe that the symmetry properties of trigonometric functions can be exploited to show that the integrals in the last line both equal $0$. $\endgroup$ – Connor Harris Jul 5 '18 at 18:26
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    $\begingroup$ Some numerical experiments seem to show that $\int_0^\pi \cos (2^n x + k \sin (3x))\, dx = 0$ for any nonnegative integer $n$ and any real $k$. $\endgroup$ – Connor Harris Jul 5 '18 at 21:25

Before we start, let us look at a related family of integrals.
For any integer $n$ and $\lambda \in \mathbb{R}$, let $J_n(\lambda)$ be the integral

$$J_n(\lambda) \stackrel{def}{=} \int_{-\pi}^{\pi} e^{in(x+\lambda\sin(3x))} dx$$

It is easy to see $J_0(\lambda) = 2\pi$ independent of $\lambda$. Furthermore, $J_n(\lambda) = 0$ unless $3$ divides $n$.

To see this, we use the fact $\sin(3x)$ is periodic with period $\frac{2\pi}{3}$. This allows us to rewrite $J_n(\lambda)$ as $$\left(\int_{-\pi}^{-\frac{\pi}{3}} + \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} + \int_{\frac{\pi}{3}}^{\pi}\right)e^{in(x+\lambda\sin(3x))} dx = \left(\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}e^{in(x+\lambda\sin(3x))} dx\right) \left(e^{-i\frac{2\pi n}{3}} + 1 + e^{i\frac{2\pi n}{3}}\right) $$ When $n$ is not divisible by $3$, $J_n(\lambda)$ vanishes because of the factor $e^{-i\frac{2\pi n}{3}} + 1 + e^{i\frac{2\pi n}{3}}$.

Back to the original problem. Since both $x$ and $\sin(3x)$ is an odd function, so does the sum. Together with $\sin^4(x)$ is an even function, we find the integrand is an even function.
As a result,

$$\begin{align}\int_0^\pi \sin^4(x + \sin(3x)) dx &= \frac12\int_{-\pi}^\pi \sin^4(x + \sin(3x))dx\\ &= \frac12\int_{-\pi}^\pi\left(\frac{ e^{i(x+\sin(3x))} - e^{-i(x+\sin(3x))}}{2i}\right)^4 dx\\ &= \frac{1}{32}\left[ J_4(1) - 4 J_2(1) + 6J_0(1) - 4J_{-2}(1) + J_{-4}(1)\right]\\ &= \frac{1}{32}\left[ 0 - 4(0) + 6(2\pi) - 4(0) + 0\right]\\ &= \frac{3\pi}{8} \end{align} $$

About the family of integrals mentioned in question/comment, we have

$$\int_0^\pi \cos (2^n x + k \sin (3x)) dx = \frac12 \int_{-\pi}^\pi \cos (2^n x + k \sin (3x)) dx = \frac14 \left(J_{2^n}(k') + J_{-2^n}(k')\right) $$ where $k' = \frac{k}{2^n}$. Since $2^n$ is not divisible by $3$, all of them evaluate to $0$.

  • $\begingroup$ Brilliant! I was trying to evaluate $\int_0^\pi \cos(2^n x + k \sin (3x)), dx$ by working out some general formula for $\int_0^\pi \cos (2f(x)\, dx)$ in terms of $\int_0^\pi \cos f(x)\, dx$, and failing badly. $\endgroup$ – Connor Harris Jul 6 '18 at 13:33

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