How do you integrate [sin(3x)1+cos(3x)+x3e1−x4]dx?

1 Answer
mason m
Jun 12, 2016

−13ln(|1+cos(3x)|)−14e1−x4+C

Explanation:

We have:

∫[sin(3x)1+cos(3x)+x3e1−x4]dx

Split this into two integrals:

=∫sin(3x)1+cos(3x)dx+∫x3e1−x4dx

Examining just the first integral:

Let A=∫sin(3x)1+cos(3x)dx.

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

A=13∫sin(3x)1+cos(3x)(3)dx=13∫sin(u)1+cos(u)du

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

A=−13∫−sin(u)1+cos(u)du=−13∫dvv

Note that ∫dvv=ln(|v|)+C.

A=−13ln(|v|)+C=−13ln(|1+cos(u)|)+C

A=−13ln(|1+cos(3x)|)+C

Now, onto the second integral:

Let B=∫x3e1−x4dx.

Again, we will use substitution: let t=1−x4, implying that dt=−4x3dx.

B=−14∫−4x3e1−x4dx=−14∫etdt

Note that ∫etdt=et+C.

B=−14et+C=−14e1−x4+C

Combining A and B, the complete integral is:

−13ln(|1+cos(3x)|)−14e1−x4+C

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