Indefinite Integral

1. Dec 6, 2009

chimychang

1. The problem statement, all variables and given/known data

$$\int{ x^3 \sqrt{(36-x^2)}dx}$$

2. Relevant equations

3. The attempt at a solution

I tried using trig substitution but got $$7776\int{cos^3(\theta)-cos^5(\theta)d\theta}$$ which seems completely wrong

$$6cos(\theta)=x$$
$$6sin(\theta)d\theta=dx$$
$$6sin(\theta)=\sqrt{36-x^2}$$

$$\int{ (6cos(\theta))^36sin(\theta)6sin(\theta)d\theta$$

2. Dec 6, 2009

HallsofIvy

Staff Emeritus
You have an odd number of "x"s outside the square root so try this: Write the integral as $\int x^2\sqrt{36- x^2}(x dx)$ and let u= 36- x2. Then du= 2x dx so x dx= (1/2)du and $x^2= 36- u$.