$\int_{-1}^{+1}\sin(\sqrt{1-x^2})\cos(x)\,dx$ =?

Question

I am trying to find the value of the definite integral: $$\int_{-1}^{+1}\sin\left(\sqrt{1-x^2}\right)\cos(x)\,dx$$

The answer from WolframAlpha is $1.20949$ but I can't solve it analytically.

Answer 1

$$\begin{eqnarray*}I=\int_{-1}^{1}\sin\left(\sqrt{1-x^2}\right)\cos(x)\,dx &=& 2\int_{0}^{1}\sin\left(\sqrt{1-x^2}\right)\cos(x)\,dx\\ &=& 2\int_{0}^{\pi/2}\sin(\cos\theta)\cos(\sin\theta)\cos(\theta)\,d\theta\end{eqnarray*} $$ can be computed in closed form by recalling that: $$ \sin(\cos\theta) = 2\sum_{k\geq 0}(-1)^k J_{2k+1}(1)\cos((2k+1)\theta), $$ $$ \cos(\sin\theta) = J_0(1)+2\sum_{k\geq 1} J_{2k}(1)\cos(2k\theta),$$ with $J_n$ being a Bessel function of the first kind, are very fast-converging Fourier series, and $$ \int_{0}^{\pi/2}\cos((2m+1)\theta)\cos(\theta)\cos(2n\theta)\,d\theta $$ for $m,n\in\mathbb{N}$, differs from zero iff $m=n$, and in such a case equals $\frac{\pi}{8}$. That gives:

$$ I = \pi \sum_{k\geq 0}(-1)^k \left[J_{2k}(1)\cdot J_{2k+1}(1)+J_{2k+1}(1)\cdot J_{2k+2}(1)\right]$$ that simplifies to:

$$ I = 2\pi \sum_{k\geq 0}(-1)^k\, (2k+1)\, J_{2k+1}(1)^2. $$

The last series converges extremely fast to the value of $I$, namely $1.209492157117477668244952433\ldots$, since $J_n(1)$ behaves like $\frac{1}{2^n n!}$.

Answer 2

$\int_{-1}^1\sin\sqrt{1-x^2}\cos x~dx$

$=2\int_0^1\sin\sqrt{1-x^2}\cos x~dx$

$=2\int_0^\frac{\pi}{2}\sin\cos\theta\cos\sin\theta\cos\theta~d\theta$

$=\int_0^\frac{\pi}{2}\sin(\cos\theta+\sin\theta)\cos\theta~d\theta+\int_0^\frac{\pi}{2}\sin(\cos\theta-\sin\theta)\cos\theta~d\theta$

$=\int_0^\frac{\pi}{2}\sin\left(\sqrt2\cos\left(\theta-\dfrac{\pi}{4}\right)\right)\cos\theta~d\theta-\int_0^\frac{\pi}{2}\sin\left(\sqrt2\sin\left(\theta-\dfrac{\pi}{4}\right)\right)\cos\theta~d\theta$

$=\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\cos\theta)\cos\left(\theta+\dfrac{\pi}{4}\right)d\theta-\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\cos\left(\theta+\dfrac{\pi}{4}\right)d\theta$

$=\dfrac{1}{\sqrt2}\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\cos\theta)\cos\theta~d\theta-\dfrac{1}{\sqrt2}\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\cos\theta)\sin\theta~d\theta-\dfrac{1}{\sqrt2}\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\cos\theta~d\theta+\dfrac{1}{\sqrt2}\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\sin\theta~d\theta$

$=\sqrt2\int_0^\frac{\pi}{4}\sin(\sqrt2\cos\theta)\cos\theta~d\theta+\sqrt2\int_0^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\sin\theta~d\theta$

$=\sqrt2\int_\frac{\pi}{2}^\frac{\pi}{4}\sin\left(\sqrt2\cos\left(\dfrac{\pi}{2}-\theta\right)\right)\cos\left(\dfrac{\pi}{2}-\theta\right)d\left(\dfrac{\pi}{2}-\theta\right)+\sqrt2\int_0^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\sin\theta~d\theta$

$=\sqrt2\int_\frac{\pi}{4}^\frac{\pi}{2}\sin(\sqrt2\sin\theta)\sin\theta~d\theta+\sqrt2\int_0^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\sin\theta~d\theta$

$=\sqrt2\int_0^\frac{\pi}{2}\sin(\sqrt2\sin\theta)\sin\theta~d\theta$

$=\dfrac{1}{\sqrt2}\int_0^\frac{\pi}{2}\cos(\theta-\sqrt2\sin\theta)~d\theta-\dfrac{1}{\sqrt2}\int_0^\frac{\pi}{2}\cos(\theta+\sqrt2\sin\theta)~d\theta$

$=\dfrac{1}{\sqrt2}\int_0^\frac{\pi}{2}\cos(\theta-\sqrt2\sin\theta)~d\theta-\dfrac{1}{\sqrt2}\int_\pi^\frac{\pi}{2}\cos(\pi-\theta+\sqrt2\sin(\pi-\theta))~d(\pi-\theta)$

$=\dfrac{1}{\sqrt2}\int_0^\frac{\pi}{2}\cos(\theta-\sqrt2\sin\theta)~d\theta+\dfrac{1}{\sqrt2}\int_\frac{\pi}{2}^\pi\cos(\theta-\sqrt2\sin\theta)~d\theta$

$=\dfrac{1}{\sqrt2}\int_0^\pi\cos(\theta-\sqrt2\sin\theta)~d\theta$

$=\dfrac{\pi J_1(\sqrt2)}{\sqrt2}$