Evaluate the integral (trig substitution)

[goosefraba]

I need help with the following problem please:

integral of: x^3 (sqrt(49-x^2)) dx

Any help would be greatly appreciated. Thanks

 

[galactus]

I suppose by your title that you have to use trig sub.

In that event, let $\displaystyle x=7sin{\theta}, \;\ dx=7cos{\theta}d{\theta}$

Make the subs and simplify it down.

$\displaystyle \int(7sin{\theta})^{3}\sqrt{49-(7sin{\theta})^{2}}\cdot 7cos{\theta}d{\theta}$

$\displaystyle \int 343sin^{3}{\theta}\sqrt{49(1-sin^{2}{\theta})}\cdot 7 cos{\theta}d{\theta}$

$\displaystyle 16807\int sin^{3}{\theta}cos^{2}{\theta}d{\theta}$

Split off a factor of sin and apply the identity $\displaystyle sin^{2}{\theta}=1-cos^{2}{\theta}$