How do you integrate #x^3/sqrt(144-x^2)#?

Question

Emma Aldridge · Accepted Answer

Use substitution with #u = 144-x^2#.

#int x^3/sqrt(144-x^2) dx# Let #u = 144-x^2#, so that #du = -2xdx# and #x^2 = 144-u# #int x^3/sqrt(144-x^2) dx = int x^2/sqrt(144-x^2)" " xdx# # = -1/2 int x^2/sqrt(144-x^2)" "(-2 x)dx# # = -1/2 int (144- u)/sqrtu" "du# # = -1/2 int (144/sqrtu - sqrtu)" "du# # = -1/2 int (144u^(-1/2) -u^(1/2))" "du# # = -1/2[288u^(1/2)-2/3u^(3/2)] +C# # = -144(144-x^2)^(1/2)+ 1/3(144-x^2)^(3/2) +C# # = -144sqrt(144-x^2)+ 1/3(sqrt(144-x^2))^3 +C#

William Abdalla · Answer

undefined To integrate $ \int \frac{x^3}{\sqrt{144-x^2}} \, dx $, you can use a trigonometric substitution. Let $ x = 12\sin(\theta) $, which implies $ dx = 12\cos(\theta) \, d\theta $. Note that $ \sqrt{144-x^2} = \sqrt{144-144\sin^2(\theta)} = 12\cos(\theta) $.

Substituting these into the integral gives:

$$ \int \frac{(12\sin(\theta))^3}{12\cos(\theta)} \cdot 12\cos(\theta) \, d\theta = \int 12^3 \sin^3(\theta) \, d\theta $$

$$ = 1728 \int \sin^3(\theta) \, d\theta $$

To integrate $ \sin^3(\theta) $, use the identity $ \sin^2(\theta) = 1 - \cos^2(\theta) $ to rewrite the integrand:

$$ = 1728 \int \sin(\theta)(1 - \cos^2(\theta)) \, d\theta $$

Let $ u = \cos(\theta) $, hence $ du = -\sin(\theta) \, d\theta $. Therefore, the integral becomes:

$$ = -1728 \int (1 - u^2) \, du $$

$$ = -1728 \left( u - \frac{u^3}{3} \right) + C $$

$$ = -1728 \left( \cos(\theta) - \frac{\cos^3(\theta)}{3} \right) + C $$

Now, substitute back $ \cos(\theta) = \frac{\sqrt{144-x^2}}{12} $ to get:

$$ = -1728 \left( \frac{\sqrt{144-x^2}}{12} - \frac{(\sqrt{144-x^2})^3}{3 \cdot 12^3} \right) + C $$

$$ = -144\sqrt{144-x^2} + \frac{\sqrt{144-x^2}^3}{3 \cdot 12^2} + C $$

$$ = -144\sqrt{144-x^2} + \frac{(144-x^2)^{\frac{3}{2}}}{288} + C $$

Thus, the integral $ \int \frac{x^3}{\sqrt{144-x^2}} \, dx $ simplifies to $ -144\sqrt{144-x^2} + \frac{(144-x^2)^{\frac{3}{2}}}{288} + C $.