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I've been wondering, what is the general form for the integral of $\sin(\cos x)$? I tried integrating it myself and couldn't, tried with Wolfram Alpha and it couldn't either. If you think about it conceptually, it should have a general form as it's graph if just like sin and cos, which both have general forms. If it doesn't have a general form, I'd like to know why. Thanks!

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    $\begingroup$ What's the antiderivative of $e^{-x^2}$? Both $e^x$ and $x^2$ have nice integrals, so of course their composition does... right? What about $\frac{\sin(x)}{x}$? What about $\frac{e^{-t}}{t}$? Welcome to the strange and wonderful world of special functions, where none of your integrals are as nice as you'd like them to be. $\endgroup$ – Omnomnomnom Dec 1 '14 at 20:29
  • $\begingroup$ en.wikipedia.org/wiki/Risch_algorithm $\endgroup$ – Count Iblis Dec 1 '14 at 21:01
  • $\begingroup$ Not every function has a "nice" anti-derivative. The antiderivatice of $\sin (\cos x)$ does exist but it cannot be written in terms of elementary function. Generally if Wolfram Alpha can't solve it then it doesn't have a general form. $\endgroup$ – Dylan Dec 1 '14 at 21:27
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What is the integral of $\sin(\cos x)$ ?

So glad you asked ! :-) Although the indefinite integral does not possess a closed form, its definite counterpart can be expressed in terms of certain special functions, such as Struve H and Bessel J.

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

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

Also, $$\int_0^\tfrac\pi2\cos(\tan x)~dx=\int_0^\tfrac\pi2\cos(\cot x)~dx=\frac\pi{2e}$$

Hope this helps ! :-)

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  • $\begingroup$ More information on this topic can be found here. $\endgroup$ – Lucian Jan 28 '16 at 17:28
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It does not have an elementary antiderivative. Calculus classes can give you an erroneous impression that you can always write down a formula for an antiderivative of a "nice" function. This is not the case.

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You can also try this one if you want.

Expanding $\sin{\cos {x}}$ in Taylor series expansion.

$\sin{\cos {x}}$ = $\cos {x}$ - $\cos^3 {x}$/3! + $\cos^5 {x}$/5! - $\cos^7 {x}$/7!+ -.....

Then one can integrate term by term.

There is no closed form solution exists for this function as Lucian suggested.

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$\int\sin(\cos x)~dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\cos^{2n+1}x}{(2n+1)!}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n+1)!}d(\sin x)$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(1-\sin^2x)^n}{(2n+1)!}d(\sin x)$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^k\sin^{2k}x}{(2n+1)!}d(\sin x)$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!\sin^{2k}x}{(2n+1)!k!(n-k)!}d(\sin x)$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!\sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$

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