How do you integrate #int sqrt(9-x^2)dx# using trigonometric substitution?

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How do you integrate #int sqrt(9-x^2)dx# using trigonometric substitution?

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Answer 1 · 2018-05-26T17:50:55

#1/2*x*sqrt(9-x^2)+9/2*arcsin(1/3*x)+C#

Explanation:

Substituting
#x=3sin(t)#
#dx=3cos(t)dt#
so we get
#9intsqrt(1-sin^2(t))cos(t)dt#
and we get
#9intcos^2(t)dt#
With #1/2*(cos(2t)+1)=cos^2(t)#
we get
#9/2int(cos(2t)+1)dt#
This is
#9/2(sin(2t)/2+t)+C#
Backsubstitution gives
#1/2xsqrt(9-x^2)+9/2arcsin(1/3x)+C#

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How do you integrate #int sqrt(9-x^2)dx# using trigonometric substitution?

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