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sin-x-3-c-dx-sinh-x-3-c-dx-




Question Number 75089 by MJS last updated on 07/Dec/19
∫sin (x^3 +c) dx=?  ∫sinh (x^3 +c) dx=?
sin(x3+c)dx=?sinh(x3+c)dx=?
Commented by mathmax by abdo last updated on 07/Dec/19
∫ sin(x^3 +c)dx =Im(∫  e^(i(x^3 +c)) dx =) and  ∫ e^(i(x^3 +c)) dx =e^(ic)  ∫ e^(i x^3 ) dx =e^(ic)  ∫  e^(−(−i x^3 )) dx  changement  −i x^3  =t  give x^3 =it ⇒ x=(it)^(1/3)  ⇒  ∫ e^(i(x^3 +c)) dx =(1/3)e^(ic)  ∫ e^(−t)  (it)^((1/3)−1)  dt  =(1/3) e^(ic)  i^((1/3)−1)  ∫  t^((1/3)−1)  e^(−t)  dt  =(1/3) e^(ic)  (e^((iπ)/2) )^(−(2/3))  ∫  t^((1/3)−1)  e^(−t)  dt  =(1/3) e^(ic−3iπ)    Γ_(in) ((1/3))  =(1/3) e^(i(c−3π))  Γ_(in) ((1/3))  Γ_(in)   is the incomplet Γ....
sin(x3+c)dx=Im(ei(x3+c)dx=)andei(x3+c)dx=eiceix3dx=eice(ix3)dxchangementix3=tgivex3=itx=(it)13ei(x3+c)dx=13eicet(it)131dt=13eici131t131etdt=13eic(eiπ2)23t131etdt=13eic3iπΓin(13)=13ei(c3π)Γin(13)ΓinistheincompletΓ.
Commented by mathmax by abdo last updated on 07/Dec/19
forgive  ∫ e^(i(x^3 +c)) dx =(1/3) e^(ic−i(π/3))    Γ_(in) ((1/3))  ⇒  I =(1/3)Γ_(in) ((1/3))sin(c−(π/3))
forgiveei(x3+c)dx=13eiciπ3Γin(13)I=13Γin(13)sin(cπ3)
Commented by MJS last updated on 07/Dec/19
thank you
thankyou
Commented by mathmax by abdo last updated on 14/Dec/19
you are welcome sir.
youarewelcomesir.

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