How to Solve the Indefinite Integral with Trigonometric Substitution?

[chimychang]

Homework Statement

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

Homework Equations

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



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

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

 

[HallsofIvy]

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 $x^2= 36- u$.