Integration - Maybe a trig identity?

[wswright]

A friend and I hvae been working on this problem for a while now, and we keep coming up with nothing. Could you possibly give us a hint as to where to start? We are supposed to integrate this:


(x^3)/sqrt(16-x^2)

 

[galactus]

Let \(\displaystyle \L\\x=4sin({\theta})\;\ and \;\ dx=4cos({\theta})\)

 

[wswright]

Thanks

Thank you! We solved the problem after seeing that hint, but we are still kind of lost as to exactly where you got 4*sin(theta) from. Could you enlighten us?

 

[stapel]

You want something you can take out of the square root. When you've got a subtraction, that usually means you want sines and cosines. Since the Pythagorean Identity is "sin²(t) + cos²(t) = 1", not "=16", you need something from which you can factor out the 16.

Eliz.

 

[skeeter]

\(\displaystyle \L \int \frac{x^3}{\sqrt{16-x^2} dx\)

\(\displaystyle \L -\int x^2 \frac{-x}{\sqrt{16-x^2}} dx\)

integration by parts ...

\(\displaystyle \L u = x^2\)

\(\displaystyle \L du = 2x dx\)

\(\displaystyle \L dv = \frac{-x}{\sqrt{16-x^2}} dx\)

\(\displaystyle \L v = \sqrt{16-x^2}\)

\(\displaystyle \L -[x^2 \sqrt{16-x^2} - \int 2x \sqrt{16-x^2} dx]\)

\(\displaystyle \L -[x^2 \sqrt{16-x^2} + \frac{2}{3}(16-x^2)^{\frac{3}{2}} + C]\)

\(\displaystyle \L -[\sqrt{16-x^2}[x^2 + \frac{2}{3}(16-x^2)] + C]\)

\(\displaystyle \L -\frac{\sqrt{16-x^2}(x^2+32)}{3} + C\)