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I am trying to calculate the indefinite integral for $0\le t\le T$ $$\int\tanh(a\sin(\omega t+\theta))dt.$$ I've tried using the integral $$\int_0^t f(\phi(\tau))d\tau=\int_0^t f^*(\tau)\phi'(\tau)d\tau=\int_{\phi(0)}^{\phi(t)}f^*(\tau)d\tau$$ where I have defined $f^*(t)=\frac{f(t)}{\phi'(t)}$. My integral (according to this) becomes $$\int\tanh(a\sin(\omega t+\theta))dt=\frac{1}{a\omega}\int_{a\sin\theta}^{a\sin(\omega t+\theta)}\tanh(\tau)\sec(\omega\tau+\theta)d\tau.$$ I've tried solving this integral with by parts and it quickly becomes incredibly complicated, so I'm hoping there my might be a simpler way to arrive at a solution.

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It is unlikely to have a closed-form antiderivative. The Risch algorithm shows there is no elementary antiderivative.

EDIT: Here's what I mean by using Chebyshev series. Suppose you know $|a| \le 2$, so $|a \sin(\omega t+\theta)| \le 2$. The Chebyshev series of $\tanh(x)$ on $[-2,2]$ is $$ \eqalign{ &1.11794183734017483\,T_{{1}} \left( x/2 \right)\cr & - 0.188742324770586944\,T_{{3}} \left( x/2 \right)\cr & + 0.0430046759680717469\,T_{{5}} \left( x/2 \right) \cr &- 0.0101069660909574362\,T_{{7}} \left( x/2 \right) + \ldots }$$ where $T_n$ are Chebyshev polynomials of the first kind. So $$ \eqalign{\tan(a \sin(\omega t + \theta)) &= 1.11794183734017483\,T_{{1}} \left( \sin(\omega t+\theta)/2 \right) \cr &- 0.188742324770586944\,T_{{3}} \left( \sin(\omega t+\theta)/2 \right) \cr &+ 0.0430046759680717469\,T_{{5}} \left( \sin(\omega t+\theta)/2 \right) \cr &- 0.0101069660909574362\,T_{{7}} \left( \sin(\omega t+\theta)/2 \right) + \ldots }$$ and each of these terms can be integrated explicitly.

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  • $\begingroup$ I see, maybe I will have to find a way of approximating it? $\endgroup$ – WnGatRC456 Aug 21 '18 at 18:38
  • $\begingroup$ You could, for example, use a series in powers of $a$. $\endgroup$ – Robert Israel Aug 21 '18 at 18:40
  • $\begingroup$ Sorry, could you elaborate? $\endgroup$ – WnGatRC456 Aug 21 '18 at 18:42
  • $\begingroup$ $\tanh(x) = x - (1/3) x^3 + (2/15) x^5 + \ldots$, so $$\int \tanh(a \sin(\omega t + \theta))\; dt = \int \left(a \sin(\omega t + \theta) - (1/3) a^3 \sin^3(\omega t + \theta) + (2/15) a^5 \sin^5(\omega t + \theta)+ \ldots\right)\; dt$$ and each term can be integrated in closed form. $\endgroup$ – Robert Israel Aug 21 '18 at 18:51
  • $\begingroup$ I see, sorry for misunderstanding I think will work for my purposes. $\endgroup$ – WnGatRC456 Aug 21 '18 at 18:52

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