e-cos-1-x-dx-

Question Number 126230 by I want to learn more last updated on 18/Dec/20

\int e^{\cos^{- 1} (x)} dx

Answered by Olaf last updated on 18/Dec/20

F (x) = \int e^{\arccos x} d x

Let x = \cos u, d x = - \sin u d u

F (x) = - \int e^{u} \sin u d u

F (x) = - [e^{u} \sin u - \int e^{u} \cos u d u]

F (x) = - e^{u} \sin u + [e^{u} \cos u - \int e^{u} (- \sin u d u]

2 F (x) = e^{u} \cos u - e^{u} \sin u

F (x) = \frac{1}{2} e^{u} (\cos u - \sin u)

F (x) = \frac{1}{2} e^{\arccos x} (x - \sin (\arccos x))

F (x) = \frac{1}{2} e^{\arccos x} (x - \sqrt{1 - x^{2}}) (+ C)

Commented by I want to learn more last updated on 18/Dec/20

Thanks sir .

Commented by I want to learn more last updated on 20/Dec/20

Thanks sir .

Answered by mathmax by abdo last updated on 19/Dec/20

I = \int e^{arcosx} dx we do the changement arcosx = t \Rightarrow x = cost \Rightarrow

I = \int e^{t} (- sint) dt = - \int e^{t} sint dt = - Im (\int e^{t + it} dt)

\int e^{(1 + i) t} dt = \frac{1}{1 + i} e^{(1 + i) t} + c = \frac{1 - i}{2} e^{t} (cost + isint) + c

= \frac{e^{t}}{2} {cost + isint - icost + sint} \Rightarrow I = - \frac{e^{t}}{2} (sint - cost) + C

I = \frac{e^{arcosx}}{2} (x - \sqrt{1 - x^{2}}) + C

Commented by I want to learn more last updated on 20/Dec/20

Thanks sir