# Evaluate `int (3x)/sqrt(9-x^2) dx` using substitution:

### 3 Answers | Add Yours

Given `int (3x)/sqrt(9-x^2)dx` :

Let `u=9-x^2,du=-2xdx`

Then `-3/2 int(-2/3 * 3x)/sqrt(9-x^2)dx=-3/2intu^(-1/2)du`

`-3/2intu^(-1/2)du = -3/2[2u^(1/2)+C_1]`

`=-3u^(1/2)+C` Substituting for u we get:

`=-3(9-x^2)^(1/2)+C` or `-3sqrt(9-x^2)+C`

---------------------------------------------------------------

`int(3x)/sqrt(9-x^2)dx=-3sqrt(9-x^2)+C`

---------------------------------------------------------------

We can check by taking the derivative:

`d/(dx)[-3(9-x^2)^(1/2)+C]=(-3)(1/2)(9-x^2)^(-1/2)(-2x)=(3x)/sqrt(9-x^2)` as required.

Let x=3sin(t) ,dx=3cos(t)dt

sqrt(9-x^2)=3cos(t)

`int(3x)/sqrt(9-x^2)dx=int(9sin(t) 3cos(t)dt)/(3cos(t)`

`int9sin(t)dt=-9cos(t)+c`

`=-9sqrt(1-x^2/3)+c`

`=-9sqrt(9-x^2)/3+c`

`=-3sqrt(9-x^2)+c`

ans.

`int((3x)/sqrt(9-x^2))dx` using substitution...

sorry for the problem above..