Integral of a fuction

amitzr3

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Joined
Feb 18, 2019
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1
Hey, could'nt find a solution for this.

I also attached my try.

Help would be appreciated.
 

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Subhotosh Khan

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Jun 18, 2007
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Apply:
\(\displaystyle u = sin^{-1}(x)\)

\(\displaystyle \frac{d}{dx}[sin^{-1}(x)] \ = \ \dfrac{1}{\sqrt{1-x^2}}\)

so your integral becomes (1/u) du.....

That should help
 

HallsofIvy

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Jan 27, 2012
Messages
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That looks much too complicated! Whoever gave you this problem clearly expects that you have previously learned that the derivative of \(\displaystyle y= \arcsin(x)\) is \(\displaystyle y'= \frac{1}{\sqrt{1- x^2}}\) so an obvious substitution is \(\displaystyle u= \arcsin(x)\). Then \(\displaystyle du= \frac{dx}{\sqrt{1- x^2}}\) and the integral becomes \(\displaystyle \int\frac{du}{u}\) which should be easy.
 
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