Question

How do you find `int_(pi/24)^(pi/18) 36(1+3^(6sec(6x)))(sec(6x)tan(6x))dx` ?

Answer 1

`12-6sqrt2+(3^12-3^(6sqrt2))/ln3`

Explanation:

We will take this step-by-step. The first substitution we will use is:

`u=6x`
`du=6color(white).dx`

The transformation of bounds becomes:

`x=pi/18" "=>" "u=6x=6(pi/18)=pi/3`

`x=pi/24" "=>" "u=6x=6(pi/24)=pi/4`

So:

`36int_(pi//24)^(pi//18)(1+3^(6sec(6x)))(sec(6x)tan(6x))dx`

`=6int_(pi//24)^(pi//18)(1+3^(6sec(6x)))(sec(6x)tan(6x))(6color(white).dx)`

`=6int_(pi//4)^(pi//3)(1+3^(6sec(u)))(sec(u)tan(u))du`

Now, let:

`v=6sec(u)`
`dv=6sec(u)tan(u)du`

The bounds become:

`u=pi/3" "=>" "v=6sec(u)=6/cos(pi/3)=6/(1/2)=12`

`u=pi/4" "=>" "v=sec(u)=6/cos(pi/4)=6/(1/sqrt2)=6sqrt2`

Then the integral becomes:

`=int_(pi//4)^(pi//3)(1+3^(6sec(u)))(6sec(u)tan(u)du)`

`=int_(6sqrt2)^12(1+3^v)dv`

Splitting up the integral:

`=int_(6sqrt2)^12dv+int_(6sqrt2)^12 3^vdv`

We can evaluate the first integral easily:

`=v| _ (6sqrt2)^12 + int _(6sqrt2)^12 3^vdv`

`=12-6sqrt2 + int _(6sqrt2)^12 3^vdv`

For the remaining integral, note that `3=e^ln3` so `3^v=(e^ln3)^v=e^(vln3)`.

`=12-6sqrt2+int_(6sqrt2)^12e^(vln3)dv`

Performing another substitution:

`w=vln3`
`dw=ln3color(white).dv`

The bounds:

`v=12" "=>" "w=vln3=12ln3`

`v=6sqrt2" "=>" "w=vln3=6sqrt2(ln3)`

Then:

`=12-6sqrt2+1/ln3int_(6sqrt2)^12e^(vln3)(ln3color(white).dv)`

`=12-6sqrt2+1/ln3int_(6sqrt2(ln3))^(12ln3)e^wdw`

The integral of `e^w` is itself:

`=12-6sqrt2+1/ln3(e^w)|_(6sqrt2(ln3))^(12ln3)`

`=12-6sqrt2+1/ln3e^(12ln3)-1/ln3e^(6sqrt2(ln3))`

`=12-6sqrt2+1/ln3(e^ln3)^12-1/ln3(e^ln3)^(6sqrt2)`

`=12-6sqrt2+(3^12-3^(6sqrt2))/ln3`

`approx473563.931508`