# Thread: Don't get these 2 trig integration problems?

1. ## Don't get these 2 trig integration problems?

2. Integral of x^2sqrt[a^2-x^2] from the interval [-a,0]

i SOLVED THE FIRST ONE.. bUT i DO NOT UNDERSTAND THE SECOND ONE AT ALL!!!

2. Originally Posted by elpermic
1. Integral of x^3 X square root(16-4x^2)

2. Integral of x^2sqrt[a^2-x^2] from the interval [-a,0]
The first integral doesn't require trig substitution...but if you need to have to, I would recommend using the trig sub $\displaystyle x=2\sin\theta$.

Spoiler:
In case you're wondering how to do the first one without trig sub, let $\displaystyle u=16-4x^2\implies x^2=\tfrac{1}{4}(16-u)$. Therefore, $\displaystyle du=-8x\,dx \implies -\frac{1}{8}\,du=dx$.

Therefore,

$\displaystyle \displaystyle\int x^3\sqrt{16-4x^2}\,dx \xrightarrow{u=16-x^2}{}-\tfrac{1}{8}\int x^2\sqrt{u}\,du=-\tfrac{1}{32}\int (16-u)\sqrt{u}\,du$

which can be easily integrated.

For the second integral, I would consider using the trig sub $\displaystyle x=a\sin\theta$.