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I am trying to solve this integral but do not get any point to start.I was thinking an U-substitution may help but do not know what to consider as U. Tried with mathematica but it does not provide any solution. Can anyone help? $\int sin\theta *2\sin^{-1} [\frac{\cos \theta+\sin \gamma}{\sin \theta \tan \beta}] d\theta$

T. an
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  • If Mma doesn't give an answer, it's fairly likely that there is no way to reduce it to standard functions. – The_Sympathizer Jul 04 '19 at 07:55
  • @The_Sympathizer I do not completely agree with you because I did one integration by hand where mathematica failed to solve. It might be my limited knowledge on mathematica as well. I always use basic commands there. – T. an Jul 04 '19 at 08:02
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    @Tanvir: cases where you can beat Mathematica are exceptional. –  Jul 04 '19 at 08:03
  • @Tanvir : Yup, but that's why I said merely "likely", not "certainly". – The_Sympathizer Jul 04 '19 at 08:03
  • @Yves Daoust: Though, to be fair, there are more published special functions than Mma has in its repertoire and one of them might be able to ork it. E.g. there are various exotic types of hypergeometric function Mma does not implement like Kampe de Feriet, which can be used to integrate $e^{\sin(x)}$, a famously Mma-resistant integrand. – The_Sympathizer Jul 04 '19 at 08:04
  • @The_Sympathizer here is one example https://math.stackexchange.com/questions/3218069/definite-double-integral – T. an Jul 04 '19 at 08:05
  • @Tanvir : Mind letting on where this integral comes from? Perhaps there is some way to get around it. It looks like it's from a geometric application of some kind. – The_Sympathizer Jul 04 '19 at 08:06
  • @The_Sympathizer: sure sir. This may help you to understand https://math.stackexchange.com/questions/3266161/area-of-a-spherical-cap-across-an-intersection – T. an Jul 04 '19 at 08:08
  • @The_Sympathizer: this example is precisely one where you can't beat Mathematica, as a rare ad-hoc function is needed. The message from Mathematica will be "don't even try". –  Jul 04 '19 at 08:28

1 Answers1

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Hint:

With $t:=\cos\theta$, renaming the constants and dropping the $2$,

$$I:=-\int \arcsin\frac{t+c}{b\sqrt{1-t^2}}dt.$$

By parts,

$$I:=-t\arcsin\frac{t+c}{b\sqrt{1-t^2}}+\int t\frac{t(t+c)+1-t^2}{b(1-t^2)^{3/2}\sqrt{1-\dfrac{(t+c)^2}{b^2(1-t^2)}}}dt \\=\cdots+\int t\frac{ct+1}{(1-t^2)\sqrt{b^2(1-t^2)-(t+c)^2}}dt.$$

This rationalizes the integral and Alpha is able to integrate. Strange.

Update:

Actually, Alpha does the integration by parts:

https://www.wolframalpha.com/input/?i=integrate+arcsin((t%2Bc)%2F(sqrt(1-t%5E2)+b))

  • Sir, which alpha you are talking about? – T. an Jul 04 '19 at 08:32
  • @Tanvir: https://www.wolframalpha.com/input/?i=t+(ct%2B1)%2Fsqrt(b%5E2(1-t%5E2)-(c%2Bt)%5E2)%2F(1-t%5E2). –  Jul 04 '19 at 08:33
  • Yeah now it able to integrate but before it gives some result after quite long time and that was really awkward. – T. an Jul 04 '19 at 08:35
  • Now out of curiosity one question comes in my mind, why alpha was not able to use Usubstitution and by parts method? – T. an Jul 04 '19 at 08:37
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    @Tanvir: computation time exceeded. Try again with Mathematica. –  Jul 04 '19 at 08:39
  • for me it gives some awkward solution without any simplification as I said. Now the solution is much convenient. – T. an Jul 04 '19 at 08:51