Question

How do I take the integral of this?

∫(x^2)/((16-x^2)^(3/2))dx

I keep on getting the wrong answer. Please help?

Answer 1

`x/sqrt(16-x^2)-sin^-1(x/4)+C`

Explanation:

`intx^2/(16-x^2)^(3/2)dx`

Let's use the trigonometric substitution `x=4sintheta`. This implies that `dx=4costhetad theta`. Then the integral equals:

`=int(16sin^2theta)/(16-16sin^2theta)^(3/2)(4costhetad theta)`

`=64int(sin^2thetacostheta)/(16^(3/2)(1-sin^2theta)^(3/2))d theta`

Recall that `1-sin^2theta=cos^2theta`:

`=64/64int(sin^2thetacostheta)/cos^3thetad theta`

`=intsin^2theta/cos^2thetad theta`

`=inttan^2thetad theta`

Use the identity `tan^2theta=sec^2theta-1`:

`=int(sec^2theta-1)d theta`

Both of these terms can be integrated easily:

`=tantheta-theta+C`

From our substitution `x=4sintheta`, we can rearrange to find that `theta=sin^-1(x/4)`.

Also, since `sintheta=x/4`, we have a right triangle where the side opposite `theta` is `x` and the hypotenuse is `4`. From the Pythagorean Theorem, we find that the side adjacent to `theta` is `sqrt(16-x^2)`. Thus, `tantheta=x/sqrt(16-x^2)`. The integral then equals:

`=x/sqrt(16-x^2)-sin^-1(x/4)+C`