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

Use substitution with

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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 ).

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