How do you use substitution to integrate #sqrt(4-x^2) dx#?

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Answer 1 · 2015-06-23T21:02:16

This can be done from trig substitution. Notice how

#sqrt(a^2 - x^2) prop sqrt(a^2 - a^2sin^2theta) prop sqrt(4 - x^2)#
where #a = 2#

so let:
#x = 2sintheta#
#dx = 2costhetad theta#
#sqrt(4-x^2) = 2costheta#

#=> int 2costheta*2costhetad theta#

#= 4int cos^2thetad theta#

Now you can use the identity:
#cos^2theta = (1+cos(2theta))/2#

Thus:
#= 2int d theta + 2int cos(2theta)d theta#

#= 2int d theta + 2*1/2int 2cos(2theta)d theta#

#= 2theta + sin(2theta) + C#

Since #x = 2sintheta#, #theta = arcsin(x/2)#.
Since #sin(2theta) = 2sinthetacostheta#:

#sin(2theta) = (xsqrt(4-x^2))/2#

#=> color(blue)(2arcsin(x/2) + (xsqrt(4-x^2))/2 + C)#