How do you integrate #intsqrt(1-4x^2)# by trigonometric substitution?

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Answer 1 · 2016-10-10T16:30:08

#sin^2(theta)/2+c#

As shown in the picture below our equation matches the "#sin#" trigonometric substitution.
enter image source here

Therefor,

#x=sin(theta)/2#

#sqrt(1-4x^2)=cos(theta)#

#dx=(2cos(theta))/4 d theta#=#cos(theta)/4#

#intcos(theta)(cos(theta))/2#=#1/2intcos^2(theta)d theta#

#1/2sin^2(theta)+c# or #sin^2(theta)/2+c#

Good Luck!!!