Challenging Indefinite Integral e^arccos x

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SUMMARY

The integral of the function \((e^{\arccos(x)}) \frac{dx}{\sqrt{1-x^2}}\) can be solved using substitution methods. The initial substitution \(u = \arccos(x)\) leads to the transformation of the integral into a more manageable form. The user struggled with further substitutions but ultimately realized the solution lies in applying trigonometric identities, specifically \( \cos^2(u) + \sin^2(u) = 1\). The Maple 11 program successfully integrates the function, confirming the validity of the approach.

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Homework Statement



Integrate:
((e^arccos(x))(dx))/(sqrt(1-x^2))

Image of the question is attached.

Homework Equations





The Attempt at a Solution



I think i took the right first step for substitution:

u=arccos(x);
cos(u)=x;
(du)(-sin(u))=dx

Substituting u into the equation:
((e^u)(-sin(u))(du))/(sqrt(1-(cos(u))^2))

This is where I'm stuck.
I tried all of the following second substitutions, but they either took me in circles or didn't replace all the u variables, or took me back to square one:

v=cos(u)
v=e^u
v=1-(cos(u))^2

I am doing this for a Calculus project using the Maple 11 program. It seems to be able to integrate the problem just fine, but I can't figure it out.
A .bmp image is attached if you are unsure of what I have typed for the question.
 

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d/dx(arcosx) is not what you have. Recheck that.


EDIT: nvm that.

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Last edited:
cos^2(u) + sin^2(u) = ...
 
Count Iblis said:
cos^2(u) + sin^2(u) = ...

Wow. I would say I can't believe how simple the question truly was, but trigonometric identities always get me. Thanks for pointing that out to me. I appreciate the help.
 

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