Question

What is `int tan^3(3x)sec^4(3x)dx`?

Answer 1

`1/36tan^4 3x(2tan^2 3x+3)+C, or,`

`1/36tan^4 3x(2sec^2 3x+1)+C`.

Explanation:

Suppose that, `I=inttan^3 3xsec^4 3xdx`,

`=inttan^3 3xsec^2 3x sec^2 3xdx`,

`=inttan^3 3x(tan^2 3x+1)sec^2 3xdx`,

`=int(tan^5 3x+tan^3 3x)sec^2 3xdx`.

Now, we subst. `tan3x=y," so that, "(sec^2 3x)(3)dx=dy`.

`:. I=1/3int(tan^5 3x+tan^3 3x)3sec^2 3xdx`.

`=1/3int(y^5+y^3)dy`,

`=1/3{y^6/6+y^4/4}`,

`=1/3*y^4/12(2y^2+3)`.

`rArr I=1/36tan^4 3x(2tan^2 3x+3)+C, or,`

`I=1/36tan^4 3x(2sec^2 3x+1)+C`.

Enjoy Maths.!

Answer 2

The answer is `=1/12tan^4(3x)+1/18tan^6(3x)+C`

Explanation:

Perform the substitition

Let `u=3x`, `=>`, `du=3dx`

Therefore,

`I=inttan^3(3x)sec^4(3x)dx=1/3inttan^3usec^4udu`

`sec^2u=1+tan^2u`

So,

`I=1/3inttan^3usec^2u(1+tan^2u)du`

Let `v=tanu`, `=>`, `dv=sec^2udu`

`I=1/3intv^3(1+v^2)dv`

`=1/3intv^3dv+1/3intv^5dv`

`=1/12v^4+1/18v^6`

`=1/12tan^4u+1/18tan^6u`

`=1/12tan^4(3x)+1/18tan^6(3x)+C`