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

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Answer 1 · 2016-06-12T16:53:19

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

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:

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

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