Question

How do you integrate `[sin(3x)/(1+cos(3x)] + x^3e^(1-x^4)] dx`?

Answer 1

`-1/3ln(abs(1+cos(3x)))-1/4e^(1-x^4)+C`

Explanation:

We have:

`int [sin(3x)/(1+cos(3x)] + x^3e^(1-x^4)] dx`

Split this into two integrals:

`=intsin(3x)/(1+cos(3x))dx+intx^3e^(1-x^4)dx`

Examining just the first integral:

Let `A=intsin(3x)/(1+cos(3x))dx`.

We can use substitution: let `u=3x`, which implies that `du=3dx`.

`A=1/3intsin(3x)/(1+cos(3x))(3)dx=1/3intsin(u)/(1+cos(u))du`

We can now use substitution again: let `v=1+cos(u)`, which implies that `dv=-sin(u)du`.

`A=-1/3int(-sin(u))/(1+cos(u))du=-1/3int(dv)/v`

Note that `int(dv)/v=ln(absv)+C`.

`A=-1/3ln(absv)+C=-1/3ln(abs(1+cos(u)))+C`

`A=-1/3ln(abs(1+cos(3x)))+C`

Now, onto the second integral:

Let `B=intx^3e^(1-x^4)dx`.

Again, we will use substitution: let `t=1-x^4`, implying that `dt=-4x^3dx`.

`B=-1/4int-4x^3e^(1-x^4)dx=-1/4inte^tdt`

Note that `inte^tdt=e^t+C`.

`B=-1/4e^t+C=-1/4e^(1-x^4)+C`

Combining `A` and `B`, the complete integral is:

`-1/3ln(abs(1+cos(3x)))-1/4e^(1-x^4)+C`