Evaluate the integral $\displaystyle\int_0^\pi \sin^4\left(x+\sin 3x\right)dx$
My work:
Let $I=\int_0^\pi \sin^4\left(x+\sin 3x\right)dx$
$=\int_0^\pi \frac18\left(\cos (4x+4\sin 3x)-4\cos(2x+2\sin 3x)+3\right)dx$
$=\frac{3\pi}{8}+\frac18\int_0^\pi\cos (4x+4\sin 3x)dx-\frac12\int_0^\pi\cos (2x+2\sin 3x)dx$
On
$$I=\frac{3\pi}{8}+\frac18\int_0^\pi\cos (4x+4\sin 3x)dx-\frac12\int_0^\pi\cos (2x+2\sin 3x)dx$$
A Simple Approach: Residue Method
Since both of above integrands are even, it is equivalent to show the following vanish.
$$I_1=\int_{-\pi}^\pi\cos (4x+4\sin 3x)dx=0,~~~~~ I_2=\int_{-\pi}^\pi\cos (2x+2\sin 3x)dx=0$$
Let $z=e^{ix},$ we get
$$I_1=\Re\oint \frac{z^3}ie^{2z^3}e^{-\frac{2}{z^{3}}}~dz,~~~~~~~I_2=\Re\oint \frac{z}ie^{z^3}e^{-\frac{1}{z^{3}}}~dz$$
Next, we inspect the coefficient for the $z^{-1}$ term in their Laurent series:
$$\begin{align}\frac{z^3}ie^{2z^3}e^{-\frac{2}{z^{3}}}&=\frac{1}{i}\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{(-1)^n2^{m+n}}{m!n!}z^{3(m-n)+3}\\ \\ \frac{z}ie^{z^3}e^{-\frac{1}{z^{3}}}&=\frac{1}{i}\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{(-1)^n}{m!n!}z^{3(m-n)+1}\end{align}$$
Neither of them have the $z^{-1}$ term in their Laurent series, hence both of their residues are zero, namely, $$I_1=I_2=0$$
Therefore,
$$\boxed{\int_0^\pi \sin^4\left(x+\sin 3x\right)dx=\frac{3\pi}8}$$
On
$$I=\frac{3\pi}{8}+\frac18\underbrace{\int_0^\pi\cos (4x+4\sin 3x)dx}_{I_1}-\frac12\underbrace{\int_0^\pi\cos (2x+2\sin 3x)dx}_{I_2}$$
A Simple Approach: Series Method
$$\cos (4x+4\sin 3x)=\cos(4x)\cos(4\sin 3x)-\sin(4x)\sin(4\sin 3x)$$
Take the Maclaurin series expansion:
$$\cos(4\sin 3x)=\sum_{n=0}^\infty a_n\sin^{2n}(3x),~~~~~\sin(4\sin 3x)=\sum_{n=0}^\infty b_n\sin^{2n+1}(3x)$$ Use the property of orthogonality for the inner product $\langle\cos (4x),\sin^{2n}(3x)\rangle=0$ and $\langle\sin(4x),\sin^{2n+1}(3x)\rangle=0$ on $[0, \pi]$, we get
$$\int_0^\pi\cos (4x)\sin^{2n}(3x)dx=0,~~~~~\int_0^\pi\sin(4x)\sin^{2n+1}(3x)dx=0$$
Namely, the integral $I_1=0$. Using the same method, we have $I_2=0$.
Therefore, $$\boxed{\int_0^\pi \sin^4\left(x+\sin 3x\right)dx=\frac{3\pi}8}$$
On
I post it, becouse my solution semms easier.
Let $\phi(x)=x+\sin(3x)$ and $I_0=\int_{0}^{\pi}dx \sin^4(\phi(x))$
$$ 2\pi=\int_{-\pi}^{\pi}dx \left[\sin^2(\phi(x))+\cos^2(\phi(x))\right]^2=\int_{-\pi}^{\pi}dx \left[\sin^4(\phi(x))+\cos^4(\phi(x))+2\sin^2(\phi(x))\cos^2(\phi(x))\right]=4 I_0+\frac{1}{2}\int_{-\pi}^{\pi}dx\sin^2(2\phi(x))\\=4I_0+\frac{1}{4}\int_{-\pi}^{\pi}dx\left[\sin^2(2\phi(x))+\cos^2(2\phi(x))\right]=4I_0+\frac{\pi}{2} $$ From this identity we straithforwardly obtain the result $$I_0=\frac{3\pi}{8}$$.
On
Without series expansion:
$\begin{align} \int_0^\pi \cos(4x+4\sin3x)dx &\stackrel{x\rightarrow x+\pi}= \int_{-\pi}^0\cos(4x-4\sin3x)dx\\ &\stackrel{x\rightarrow -x} =\int_0^\pi \cos(-4x+4\sin3x)dx\\ &\;\;=\int_0^\pi\cos(4x-4\sin3x)dx\\ &\;\;=\int_0^\pi\cos4x\cos(4\sin3x)dx\\ &\stackrel{\text{IBP}}{=}3\int_0^\pi\sin4x\sin(4\sin3x)\cos3xdx\\ &=\frac32\int_0^\pi(\sin7x+\sin x)\sin(4\sin3x)dx\\ &\;\;=0 \end{align}$
Similarly, we can show that $$\int_0^\pi\cos(2x+2\sin3x)dx=0.$$ Hence, from the equation that OP found, we have $I=\frac38\pi$.
On
Since the question has already been answered, I would like to share the generalized versions of the problem:
(1) Let $n,k\in N$ such that $2p\not\equiv 0 \pmod {(2k+1)}$ $\forall p\in \{1,2,3,\ldots, n\}$ then $$\displaystyle \int_0^{\pi}\left[\sin\big(x+\sin((2k+1)x)\big)\right]^{2n}dx=\displaystyle \frac {{2n-1 \choose n}}{2^{2n-1}}\pi$$
(2) $$\displaystyle \int_0^{\pi} \left[\sin \big(x+\sin((2k+1)x)\big)\right]^{2n}dx$$ $$=\displaystyle \frac {{2n-1 \choose n}\pi}{2^{2n-1}}+\frac {\pi}{2^{2n-1}(2k+1)}\sum_{i=1}^{\left\lfloor \frac {n}{2k+1}\right\rfloor} \sum_{r=0}^{2k} (-1)^i {2n \choose n-(2k+1)i}J_{2i}(2i(2k+1)\cdot (-1)^{r+1})$$ where $J_{\nu}(z)$ denotes Bessel function of first kind.
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 $0$.