I am stuck on taking the integral of this:
$$\int \sin(9\sinh(2x))\,{\rm d}x$$
I tried using using a double $u$ substitution, but i got confused. Nothing seems to cancel out.
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1This does not look like an integral that has an elementrary antiderivative. – Winther Feb 04 '17 at 02:03
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how would i start solving it? – prince k Feb 04 '17 at 02:08
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You can likely find a solution in the form of a power-series. I might be wrong, but I won't expect this to have a simple closed form solution. – Winther Feb 04 '17 at 02:11
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See Bessel and Struve functions. – Lucian Feb 04 '17 at 02:19
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In general, $~\displaystyle\int_0^\infty\sin(a~\sinh t)~dt~=~\dfrac\pi2~\Big(I_0(a)-L_0(a)\Big),~$ where I and L represent the Bessel and Struve functions, respectively. Letting $t=2x$ and $a=9$ should therefore yield a closed form expression for the definite integral in terms of these two special functions. However, evaluating the indefinite integral would require the existence of “incomplete” Bessel and Struve functions, which, unfortunately, do not formally exist $($as a conventional object of study within the larger realm of mathematics$)$.
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$\int\sin(9\sinh2x)~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n9^{2n+1}\sinh^{2n+1}2x}{(2n+1)!}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n9^{2n+1}\sinh^{2n}2x}{2(2n+1)!}~d(\cosh2x)$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n9^{2n+1}(\cosh^22x-1)^n}{2(2n+1)!}~d(\cosh2x)$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{2n-k}9^{2n+1}C_k^n\cosh^{2k}2x}{2(2n+1)!}~d(\cosh2x)$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^k9^{2n+1}n!\cosh^{2k+1}2x}{2(2n+1)!k!(n-k)!(2k+1)}+C$
Harry Peter
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