Question

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

Answer 1

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

Explanation:

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

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

Explanation:

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)`