∫ e^( cos^(− 1) (x)) dx


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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
	$I=∫ e^(arcosx) dx we do the changement arcosx=t ⇒x=cost ⇒ I=∫ e^t (−sint)dt =−∫ e^t sint dt =−Im(∫ e^(t+it) dt) ∫ e^((1+i)t) dt =(1/(1+i))e^((1+i)t) +c=((1−i)/2)e^t (cost +isint) +c =(e^t /2){cost +isint −icost +sint} ⇒I=−(e^t /2)( sint−cost) +C I =(e^(arcosx) /2)(x−(√(1−x^2 ))) +C$
	$I = \int e^{arcosx} dx we do the changement arcosx = t \Rightarrow x = cost \Rightarrow$ $I = \int e^{t} (- sint) dt = - \int e^{t} sint dt = - Im (\int e^{t + it} dt)$ $\int e^{(1 + i) t} dt = \frac{1}{1 + i} e^{(1 + i) t} + c = \frac{1 - i}{2} e^{t} (cost + isint) + c$ $= \frac{e^{t}}{2} {cost + isint - icost + sint} \Rightarrow I = - \frac{e^{t}}{2} (sint - cost) + C$ $I = \frac{e^{arcosx}}{2} (x - \sqrt{1 - x^{2}}) + C$
	Commented by I want to learn more last updated on 20/Dec/20

	$Thanks sir$
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