# Find the volume of the solid of revolution obtained by rotating the curve x=3cos^3theta , y=3sin^3theta about the x axis? Sep 19, 2017

See a way to do the integrals involved more quickly below.

#### Explanation:

I'm not going to answer the question at hand here, but I thought it would be helpful to point out that the integrals with odd powers of sine and cosine can be solved more quickly.

For example, to do $\setminus \int \setminus {\sin}^{9} \left(\theta\right) d \setminus \theta$ quickly, it's best to write ${\sin}^{9} \left(\theta\right) = {\left({\sin}^{2} \left(\theta\right)\right)}^{4} \cdot \sin \left(\theta\right) = {\left(1 - {\cos}^{2} \left(\theta\right)\right)}^{4} \cdot \sin \left(\theta\right)$. With the substitution $u = \cos \left(\theta\right)$, $\mathrm{du} = - \sin \left(\theta\right) d \setminus \theta$, the integral then can be solved as follows (with the help of the binomial theorem):

$\int \setminus {\sin}^{9} \left(\theta\right) d \setminus \theta = \int \setminus - {\left(1 - {u}^{2}\right)}^{4} \mathrm{du}$

$= \int \left(- {u}^{8} + 4 {u}^{6} - 6 {u}^{4} + 4 {u}^{2} - 1\right) \mathrm{du}$

$= - \frac{1}{9} {u}^{9} + \frac{4}{7} {u}^{7} - \frac{6}{5} {u}^{5} + \frac{4}{3} {u}^{3} - u + C$

$= - \frac{1}{9} {\cos}^{9} \left(\theta\right) + \frac{4}{7} {\cos}^{7} \left(\theta\right) - \frac{6}{5} {\cos}^{5} \left(\theta\right) + \frac{4}{3} {\cos}^{3} \left(\theta\right) - \cos \left(\theta\right) + C$.

Integrals involving only even powers of sine and cosine are harder to solve quickly. The identities ${\cos}^{2} \left(\theta\right) = \frac{1}{2} + \frac{1}{2} \cos \left(2 \theta\right)$ and ${\sin}^{2} \left(\theta\right) = \frac{1}{2} - \frac{1}{2} \cos \left(2 \theta\right)$ are helpful, but they are still typically a lot of work.