Double substitution when integrating.

Question

I need to integrate $$f(x) = \cos(\sin x)$$ my first thought was substituting so that $u = \cos(x)$, but that doesn't seem to do the trick. Is there any way to do a double substitution on this?

Any ways on how to proceed would be appreciated.

Answer 1

Although the indefinite integral $($or anti-derivative$)$ cannot be expressed in terms of elementary functions, its definite counterpart does have a closed form in terms of the special Bessel function:

$$\int_0^\tfrac\pi2\cos(\sin x)~dx~=~\int_0^\tfrac\pi2\cos(\cos x)~dx~=~\frac\pi2J_0(1).$$

Also, $$\int_0^\tfrac\pi2\sin(\sin x)~dx~=~\int_0^\tfrac\pi2\sin(\cos x)~dx~=~\frac\pi2H_0(1),$$

where H represents the special Struve function.

Answer 2

I tried to use WolframAlpha but no standard integral found. So, you can approximate the function to an easier function.

cos(sin(x)) = a cos(bx)+c

if a = 0.25, b = 2, c = 0.75

then the approximation is almost inline with the function (I used a plotting program for this conclusion).

$$ \int f(x)dx = \int 0.25\cos(2x)+0.75\ dx \\= 0.5\sin(2x)\ +0.75x+C $$

Answer 3

Or you also can solve it numerically by expand the function (e.g. use Taylor expansion) and then integrate each of its terms (e.g. first 5 terms).

Answer 4

You can express in terms of Incomplete Bessel Functions:

Consider $$\begin{align}J_0(1,w)&=\dfrac{2}{\pi}\int_0^w\cos\cos x~dx \\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cos\cos\left(x-\dfrac{\pi}{2}\right)~d\left(x-\dfrac{\pi}{2}\right) \\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cos\sin x~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cos\sin x~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{2}{\pi}\int_\frac{\pi}{2}^0\cos\sin\left(\dfrac{\pi}{2}-x\right)~d\left(\dfrac{\pi}{2}-x\right) \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx-\dfrac{1}{\pi}\int_\pi^\frac{\pi}{2}\cos\cos(\pi-x)~d(\pi-x) \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx-\dfrac{1}{\pi}\int_\frac{\pi}{2}^\pi\cos\cos x~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\pi\cos\cos x~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-J_0(1)\end{align} $$

Then $\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx=J_0(1)+J_0(1,w)$

$\therefore\int\cos\sin x~dx=\dfrac{\pi}{2}\left(J_0(1)+J_0\left(1,x-\dfrac{\pi}{2}\right)\right)+C$