Question

How do you integrate `int x^3/sqrt(16-x^2)` by trigonometric substitution?

Answer 1

The integral is `(16 - x^2)^(3/2)/3 - 16sqrt(16 - x^2) + C`

Explanation:

Apply the substitution `x = 4sintheta`. Then `dx = 4costheta d theta`.

`=>4int (4sintheta)^3/sqrt(16 - (4sintheta)^2) * costheta d theta`

`=>4int (64sin^3theta)/sqrt(16 - 16sin^2theta) * costheta d theta`

`=>4int (64sin^3theta)/sqrt(16(1 - sin^2theta)) * costheta d theta`

`=>4int (64sin^3theta)/sqrt(16cos^2theta) * costheta d theta`

`=>4int (64sin^3theta)/(4costheta) * costheta d theta`

`=>4int (64sin^3theta)/4d theta`

`=>4int 16sin^3theta d theta`

`=>64intsintheta(sin^2theta) d theta`

`=>64intsintheta(1 - cos^2theta)d theta`

Let `u = costheta`. Then `du = -sintheta d theta` and `d theta = -(du)/sintheta`.

`=>-64intsintheta(1 - u^2) * (du)/sintheta`

`=>-64int1 - u^2 du`

`=>-64(u - 1/3u^3) + C`

`=>-64u + 64/3u^3 + C`

Reverse the substitution.

`=>-64costheta + 64/3cos^3theta + C`

Since the initial substitution was `sintheta = x/4`, then `costheta = sqrt(16 - x^2)/4`.

`=>-(64sqrt(16 - x^2))/4 + 64/3(sqrt(16 - x^2)/4)^3 + C`

`=>-16sqrt(16 - x^2) + (16 - x^2)^(3/2)/3 + C`

Hopefully this helps!

Answer 2

`intx^3/(sqrt(16-x^2)) dx=`
`1/3(16-x^2)^(3/2)-16sqrt(16-x^2)+"C"`

Explanation:

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

If we look at the denominator, it looks like a rearrangement of Pythagoras' theorem, `a^2+b^2=c^2 rarrc^2-b^2=a^2`

So the length of the hypotenuse of our triangle is `4` and the length of the opposite is `x`. This means that the length of the adjacent is `sqrt(16-x^2)`.

If we rewrite `x` using trigonometric ratios, we get `sintheta=x/4 rarrx=4sintheta`

We can also rewrite the adjacent in the same way:

`costheta=(sqrt(16-x^2))/4`

`sqrt(16-x^2)=4costheta`

If we differentiate `x`, we get:

`dx/(d theta)=4costheta`

`dx=4costhetad theta`

`intx^3/(sqrt(16-x^2))dx=intx^3(4costheta)/(4costheta)d theta=intx^3d theta`

Let's rewrite `x^3` using our `x` function:

`x=4sintheta`

`x^3=(4sintheta)^2=64sin^3theta`

Now our integral is `int64sin^3thetad theta=64 intsin^3thetad theta`.

`=64intsin^2theta sintheta d theta=64intsintheta(1-cos^2theta)d theta`

`u=costheta`

`(du)/(d theta)=-sintheta`

`d theta=-(1/sintheta)du`

`64intsintheta(1-cos^2theta)d theta=-64intcancel(sin)theta(1-u^2)cancel(1/sintheta)du`

`=-64int(1-u^2)du=-64(u-1/3u^3)+"C"`

Substitute `u` for `costheta` and `costheta` for `sqrt(16-x^2)/4`

`-64(u-1/3u^3)+"C"=-64(sqrt(16-x^2)/4-(1/3)(16-x^2)^(3/2)/64)+"C"`

`=1/3(16-x^2)^(3/2)-16sqrt(16-x^2)+"C"`

Answer 3

Contd.

Explanation:

We solve the Problem is required to be solved using trigo. substn.,