Question

How do you integrate `int x^2/sqrt(16-x^2)` by trigonometric substitution?

Answer 1

`8sin^-1(x/4)-(xsqrt(16-x^2))/2+C`

Explanation:

`I=intx^2/sqrt(16-x^2)color(red)(dx)`

Let's use the substitution `x=4sintheta`. This substitution implies that couple important things:

• `x^2=16sin^2theta`
• `sqrt(16-x^2)=sqrt(16-16sin^2theta)=4sqrt(1-sin^2theta)=4costheta`
• `dx=4costhetad theta`

Now, let's substitute these back into our integral:

`I=int(16sin^2theta)/(4costheta)(4costhetad theta)=16intsin^2thetad theta`

To solve this, I like to use the identity `sin^2theta=1/2(1-cos2theta)`. If you've never seen this before, remember that `cos2theta=cos^2theta-sin^2theta`, which implies that `cos2theta=1-2sin^2theta`, from which you can solve for `sin^2theta` to get the identity from before. Using the identity in the integral:

`I=16int1/2(1-cos2theta)d theta=8(theta-1/2sin2theta)+C`

We need to get our final answer back in terms of `x` instead of `theta`. Use `sin2theta=2sinthetacostheta`:

`I=8theta-8sinthetacostheta+C`

Recall we started with `x=4sintheta`. Thus:

• `sintheta=x/4`
• `costheta=sqrt(1-sin^2theta)=sqrt(1-x^2/16)=sqrt(16-x^2)/4`
• `theta=sin^-1(x/4)`

And finally:

`I=8sin^-1(x/4)-8(x/4)(sqrt(16-x^2)/4)+C`

`I=8sin^-1(x/4)-(xsqrt(16-x^2))/2+C`

Answer 2

`int x^2/sqrt(16-x^2) dx =8arcsin(x/4) - (x sqrt(16-x^2))/2 +C`

Explanation:

Let `x=4sint` with `t in(-pi/2,pi/2)`, `dx = 4costdt`, then:

`int x^2/sqrt(16-x^2) dx = int ((4sint)^2(4costdt))/sqrt(16-16sin^2t)`

`int x^2/sqrt(16-x^2) dx = 16int (sin^2tcostdt)/sqrt(1-sin^2t)`

For `t in(-pi/2,pi/2)` the cosine is positive, so:

`sqrt(1-sin^2t) = sqrt(cos^2t) = cost`

then:

`int x^2/sqrt(16-x^2) dx = 16int (sin^2tcostdt)/cost = 16 int sin^2tdt`

Use now the trigonometric identity:

`2sin^2alpha = 1-cos(2alpha)`

to have:

`int x^2/sqrt(16-x^2) dx = 8int (1-cos2t)dt`

and based on the linearity of the integral:

`int x^2/sqrt(16-x^2) dx = 8int dt - 4 int cos2td(2t)`

`int x^2/sqrt(16-x^2) dx = 8t - 4 sin2t +C`

`int x^2/sqrt(16-x^2) dx = 8t - 8 sintcost +C`

and undoing the substitution:

`int x^2/sqrt(16-x^2) dx =8arcsin(x/4) - 2x sqrt(1-(x/4)^2) +C`

`int x^2/sqrt(16-x^2) dx =8arcsin(x/4) - (x sqrt(16-x^2))/2 +C`