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#