Finding $\displaystyle \int^{\pi}_{0}\cos^4(x+\sin 3x)dx$

Try: From $\displaystyle \cos^4(x)=(\cos^2(x))^2=\frac{1}{4}\bigg[1+\cos 2x\bigg]^2$

$$=\frac{1}{4}+\frac{1}{8}\bigg(1+\cos^2(4x)+2\cos 4x\bigg)+\frac{\cos 2x}{2}$$


So $$\cos^4(x+\sin 3x)=\frac{7}{16}+\frac{1}{16}\cos(8x+8\sin 3x)+\frac{1}{4}\cos(4x+4\sin 3x)+\frac{1}{2}\cos(2x+3\sin 3x)$$

How can i solve $$\int^{\pi}_{0}\sin(8x+8\sin 3x)dx$$

$$\int^{\pi}_{0}\sin(4x+4\sin 3x)dx$$ Type integrals

I have seems that it is not possible in elementry way

could some help me to solve it . thanks


$\cos^4\theta$ can be written in terms of $1,\cos(2\theta),\cos(4\theta)$. Additionally the value of

$$ \int_{0}^{\pi}\cos\left(2kx+2k\sin(3x)\right)\,dx=\frac{1}{2}\text{Re}\int_{-\pi}^{\pi}e^{2k i x}\cdot e^{2ki\sin(3x)}\,dx $$ can be derived from the Jacobi-Anger expansion $$ e^{iz\sin\theta} = \sum_{n\in\mathbb{Z}} J_n(z) e^{in\theta} $$ leading to $$ e^{2ki\sin(3x)}=\sum_{n\in\mathbb{Z}}J_n(2k) e^{3nix}. $$ Since for any $a,b\in\mathbb{Z}$ we have $\int_{-\pi}^{\pi}e^{aix}e^{bix}\,dx = 2\pi\delta(a+b)$, for $k\in\mathbb{N}$ the integral $\int_{0}^{\pi}\cos\left(2kx+2k\sin(3x)\right)\,dx$ differs from zero only if $2k$ is a multiple of $3$ (i.e. if $k$ is a multiple of $3$), and in such a case it equals $\pi J_{2k/3}(2k)$ (with $J_n$ being a Bessel function of the first kind). Since neither $2$ or $4$ are multiples of three and $\cos^4(\theta)=\frac{3}{8}+\frac{1}{2}\cos(2\theta)+\frac{1}{8}\cos(4\theta)$, it follows that $$ \int_{0}^{\pi}\cos^4(x+\sin(3x))\,dx = \color{red}{\frac{3\pi}{8}}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.