Integral of a fuction

amitzr3

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Feb 18, 2019
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Hey, could'nt find a solution for this.

I also attached my try.

Help would be appreciated.
 

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Apply:
[MATH]u = sin^{-1}(x)[/MATH]
[MATH]\frac{d}{dx}[sin^{-1}(x)] \ = \ \dfrac{1}{\sqrt{1-x^2}}[/MATH]
so your integral becomes (1/u) du.....

That should help
 
That looks much too complicated! Whoever gave you this problem clearly expects that you have previously learned that the derivative of y=arcsin(x)\displaystyle y= \arcsin(x) is y=11x2\displaystyle y'= \frac{1}{\sqrt{1- x^2}} so an obvious substitution is u=arcsin(x)\displaystyle u= \arcsin(x). Then du=dx1x2\displaystyle du= \frac{dx}{\sqrt{1- x^2}} and the integral becomes duu\displaystyle \int\frac{du}{u} which should be easy.
 
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