Kim Baxter

Posted on Saturday, 01 March, 2003 - 11:28 am:

Hi... I have a calculus question in which I have to use a trigonometric substitution

$\int \frac{x^{2}}{\sqrt{9 - x^{2}}} dx$

I have substituted $u = 3 \sin θ$ which has given me $9 / 2 (x - x / 3 \cos θ)$

could someone tell me if I am on the right track or if it is right

Ian Short

Posted on Saturday, 01 March, 2003 - 01:10 pm:

Setting

x = 3 \sin θ

gives

dx = 3 \cos θ d θ

. So

$\int x^{2} / \sqrt{9 - x^{2}} dx$

$= \int (\sin^{2} θ) (3 \cos θ) / \sqrt{9 - 9 \sin^{2} θ} d θ$

$= \int (9 \sin^{2} θ) (3 \cos θ) / 3 \cos θ d θ$

$= \int 9 \sin^{2} θ d θ$ .

Is that clear?