How do you integrate #int 1/sqrt((1-(x^2)/3))dx# using trigonometric substitution?

1 Answer
mason m
Sep 7, 2016

#sqrt3arcsin(x/sqrt3)+C#

Explanation:

We'll use the substitution #x=sqrt3sin(theta)#. Thus, #dx=sqrt3cos(theta)d theta#. Substituting into the integral:

#intdx/sqrt(1-x^2/3)=int(sqrt3cos(theta)d theta)/sqrt(1-(3sin^2(theta))/3)=sqrt3int(cos(theta)d theta)/sqrt(1-sin^2(theta))#

Note that #sin^2(theta)+cos^2(theta)=1#, so we see that #cos(theta)=sqrt(1-sin^2(theta))#:

#=sqrt3int(cos(theta)d theta)/cos(theta)=sqrt3intd theta=sqrt3theta+C#

From #x=sqrt3sin(theta)# we see that #theta=arcsin(x/sqrt3)#:

#=sqrt3arcsin(x/sqrt3)+C#

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