How do you integrate #int sqrt(4-9x^2)# using trig substitutions?

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Answer 1 · 2016-12-30T12:06:03

#= 1/2 x sqrt (4- 9 x^2) + 2/3 sin^(-1) (3/2 x) + C#

One obvious thing here is to use a sub to get the integrand looking like this

# sqrt(4-4 sin^2 y) = 2 cos y#, ie using the Pythagorean identity

so we can say that #9 x^2 = 4 sin^2 y#, and the sub we are going to try is #x = 2/3 sin y, \ dx = 2/3 cos y \ dy#

The integration is then

#int 2 cos y * 2/3 cos y \ dy#

#= 4/3int cos^2 y \ dy#

use the double-angle identity #cos 2 A = 2 cos^2 A - 1#

#= 2/3int cos 2y + 1 \ dy#

#= 2/3 ( 1/2sin 2y + y ) + C#

now #sin 2y = 2 sin y cos y = 2 * 3/2 x * sqrt (1- (3/2 x)^2)#

giving

#= x sqrt (1- (3/2 x)^2) + 2/3 sin^(-1) (3/2 x) + C#

#= 1/2 x sqrt (4- 9 x^2) + 2/3 sin^(-1) (3/2 x) + C#

Answer 2 · 2016-12-30T13:14:22

#int sqrt(4-9x^2)dx = x/2 sqrt (4-9x^2) +2/3arcsin(3/2x)#

Write the integral as:

#int sqrt(4-9x^2)dx = 2intsqrt(1-(3/2x)^2)dx#

Now substitute:

#x=2/3 sint#, #dx=2/3cost#

#int sqrt(4-9x^2)dx = 4/3 int sqrt(1-sin^2t) cost dt= 4/3 int cos^2tdt#

To solve the last integral, we note that:

#cos2x = (2cos^2x-1)=> cos^2x = (cos2x+1)/2#

so we have:

#int sqrt(4-9x^2)dx = 2/3 (int cos2t dt + int dt)= 1/3(sin2t+2t)#

Substituting back x:

#sin2t= 2 sint cost = 3xsqrt (1-(3/2x)^2)#

#t=arcsin(3/2x)#

Finally:

#int sqrt(4-9x^2)dx = xsqrt (1-(3/2x)^2) +2/3arcsin(3/2x)#