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

Question

Cameron Alexander · Accepted Answer

#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)#

Christopher Agresta · Answer

#= 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#

Evelyn Agard · Answer

undefined To integrate $ \int \sqrt{4-9x^2} $ using trigonometric substitution, we can use the trigonometric identity $ \sin^2(\theta) + \cos^2(\theta) = 1 $ to express $ x $ in terms of $ \sin(\theta) $ or $ \cos(\theta) $.

Given $ \sqrt{4-9x^2} $, we notice that it resembles the form of $ \sqrt{a^2 - x^2} $, which can be simplified using a trigonometric substitution. In this case, we let $ x = \frac{2}{3} \sin(\theta) $, which implies $ dx = \frac{2}{3} \cos(\theta) d\theta $.

Substituting $ x $ and $ dx $ into the integral, we obtain:

$$ \int \sqrt{4-9x^2} \, dx = \int \sqrt{4-9\left(\frac{2}{3}\sin(\theta)\right)^2} \left(\frac{2}{3}\cos(\theta)\right) \, d\theta $$

After simplifying, we get:

$$ \int \sqrt{4-9x^2} \, dx = \int \sqrt{4-4\sin^2(\theta)} \left(\frac{2}{3}\cos(\theta)\right) \, d\theta $$

Now, $ \sqrt{4-4\sin^2(\theta)} $ can be simplified using the trigonometric identity $ \cos^2(\theta) = 1 - \sin^2(\theta) $:

$$ \sqrt{4-4\sin^2(\theta)} = \sqrt{4\cos^2(\theta)} = 2\cos(\theta) $$

Substituting this back into the integral, we have:

$$ \int \sqrt{4-9x^2} \, dx = \int 2\cos(\theta) \left(\frac{2}{3}\cos(\theta)\right) \, d\theta $$

$$ = \frac{4}{3} \int \cos^2(\theta) \, d\theta $$

Now, we can use the trigonometric identity $ \cos^2(\theta) = \frac{1+\cos(2\theta)}{2} $ to integrate:

$$ \int \cos^2(\theta) \, d\theta = \frac{1}{2} \int (1 + \cos(2\theta)) \, d\theta $$

$$ = \frac{1}{2} \left(\theta + \frac{\sin(2\theta)}{2}\right) + C $$

Finally, we substitute back $ \theta $ in terms of $ x $ to get the final result.