Question

How do you integrate `int x^3/sqrt(64+x^2)` by trigonometric substitution?

Answer 1

`I = 64/3(x^2 + 64)^(3/2) - 64sqrt(x^2 + 64) + C`

Explanation:

Use the substitution `x= 8tantheta`. Then `dx= 8sec^2theta d theta`.

`I = int (8tantheta)^3/sqrt(64 + (8tantheta)^2) * 8sec^2theta d theta`

`I = int (512tan^3theta)/sqrt(64 + 64tan^2theta) * 8sec^2theta d theta`

`I = int(512tan^3theta)/sqrt(64(1 + tan^2theta)) * 8sec^2theta d theta`

Now use `1 + tan^2theta = sec^2theta`.

`I = int(512tan^3theta)/sqrt(64sec^2theta) * 8sec^2theta d theta`

`I = int(512tan^3theta)/(8sectheta) * 8sec^2theta d theta`

`I = int512sec thetatan^3theta d theta`

`I =int512secthetatantheta(tan^2theta) d theta`

We now use `tan^2theta = sec^2theta -1`.

`I = int512secthetatantheta(sec^2theta - 1) d theta`

Now let `u = sectheta`. Then `du = secthetatantheta d theta` and `d theta = (du)/(secthetatantheta)`.

`I = int512secthetatantheta * (u^2- 1) (du)/(secthetatantheta)`

`I= int(512u^2 - 512)du`

`I = 512/3u^3 - 512u + C`

`I = 512/3sec^3theta - 512sectheta`

From our initial substitution, we know that `tantheta = x/8`. Then `sectheta = sqrt(x^2 + 64)/8`

`I = 64/3(x^2 + 64)^(3/2) - 64sqrt(x^2 + 64) + C`

Hopefully this helps!

Answer 2

`int (x^3)/(sqrt{64+x^2}) "d"x= 1/3sqrt{64+x^2}(x^2-128) + C`.

Explanation:

I know you specified trigonometric substitution but I don't see why you'd use on in this case as a more obvious one stands out to me, because we have an `x^2` with an `x^3` on the numerator.

Someone's already submitted the trigonometric substitution method so thought I may as well share.

`I = int (x^3)/(sqrt{64+x^2})"d"x`.

Let `u=x^2`, `("d"u)/("d"x) = 2x`, `"d"x = 1/(2x) "d"u`.

Then,

`I = 1/2 int x^2/(sqrt{64+x^2}) "d"u`,

`= 1/2 int u(64+u)^(-1/2) "d"u`.

At this point, I used integration by parts. Let `s = u`, and `dt = (64 + u)^(-1/2)du`. Then, for `int sdt = st - int tds`:

`I = 1/2 ( [2u(64+u)^(1/2)]-2int(64+u)^(1/2) "d"u)`

`I = u(64+u)^(1/2) - 2/3(64+u)^(3/2) + C`

Then, I factored out `1/3 sqrt(64 + x^2)` and substituted `u = x^2` back in:

`I = 1/3sqrt{64+x^2}(3x^2-2(64+x^2)) + C`

`= color(blue)(1/3sqrt{64+x^2}(x^2-128) + C)`.

Answer 3

`1/3sqrt(x^2+64)(x^2-128)+C.`

Explanation:

Here is another way to solve the Problem, without using

substitution.

`I=intx^3/sqrt(x^2+64)dx,`

`=int(x^2*x)/sqrt(x^2+64)dx,`

`=int{((x^2+64)-64)x}/sqrt(x^2+64)dx,`

`=int((x^2+64)/sqrt(x^2+64)-64/sqrt(x^2+64))xdx,`

`=int{(x^2+64)^(1/2)-64(x^2+64)^(-1/2)}{1/2d/dx(x^2+64)}dx,`

`=int(x^2+64)^(1/2){1/2d/dx(x^2+64)}dx`
`-64int(x^2+64)^(-1/2){1/2d/dx(x^2+64)}dx,`

`=1/2int(x^2+64)^(1/2)(d/dx(x^2+64))dx`
`-32int(x^2+64)^(-1/2)(d/dx(x^2+64))dx,`

`=1/2(x^2+64)^(1/2+1)/(1/2+1)-32(x^2+64)^(-1/2+1)/(-1/2+1),`

`=1/3(x^2+64)^(3/2)-64(x^2+64)^(1/2),`

`=1/3sqrt(x^2+64){(x^2+64)-192)},`

`=1/3sqrt(x^2+64)(x^2-128)+C.`

N.B.: The Final Integrals were obtained using the Rule,

`int[f(x)]^nf'(x)dx=[f(x)]^(n+1)/(n+1)+c, n ne-1.`

Enjoy Maths.!

Answer 4

After using `x=8sinhu` and `dx=8coshu` transforms, this integral became,

`int(8sinhu)^3*(8coshu)/(8coshu)*du`

`=int512(sinhu)^3*du`

`=int512(sinhu)^2*sinhu*du`

`=int512((coshu)^2-1)*sinhu*du`

`=int512(coshu)^2*sinhu*du-int512sinhu*du`

`=512/3*(coshu)^3-512coshu+C`

After using `sinhu=x/8` and `coshu=(Sqrt(x^2+64))/8` inverse transforms, solution of this integral became,

=`int(x^3)/(Sqrt(x^2+64))*dx`

=`=1/3*(x^2+64)^(3/2)-64Sqrt(x^2+64)+C`

=`=1/3*(x^2-128)*Sqrt(x^2+64)+C`