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

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embizze's profile pic

embizze | High School Teacher | (Level 2) Educator Emeritus

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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`




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.

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

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


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






dylzzz's profile pic

dylzzz | Student, Undergraduate | (Level 1) Honors

Posted on

`int((3x)/sqrt(9-x^2))dx` using substitution...
sorry for the problem above..