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

Jun 12, 2016

$- \frac{1}{3} \ln \left(\left\mid 1 + \cos \left(3 x\right) \right\mid\right) - \frac{1}{4} {e}^{1 - {x}^{4}} + C$

#### Explanation:

We have:

$\int \left[\sin \frac{3 x}{1 + \cos \left(3 x\right)} + {x}^{3} {e}^{1 - {x}^{4}}\right] \mathrm{dx}$

Split this into two integrals:

$= \int \sin \frac{3 x}{1 + \cos \left(3 x\right)} \mathrm{dx} + \int {x}^{3} {e}^{1 - {x}^{4}} \mathrm{dx}$

Examining just the first integral:

Let $A = \int \sin \frac{3 x}{1 + \cos \left(3 x\right)} \mathrm{dx}$.

We can use substitution: let $u = 3 x$, which implies that $\mathrm{du} = 3 \mathrm{dx}$.

$A = \frac{1}{3} \int \sin \frac{3 x}{1 + \cos \left(3 x\right)} \left(3\right) \mathrm{dx} = \frac{1}{3} \int \sin \frac{u}{1 + \cos \left(u\right)} \mathrm{du}$

We can now use substitution again: let $v = 1 + \cos \left(u\right)$, which implies that $\mathrm{dv} = - \sin \left(u\right) \mathrm{du}$.

$A = - \frac{1}{3} \int \frac{- \sin \left(u\right)}{1 + \cos \left(u\right)} \mathrm{du} = - \frac{1}{3} \int \frac{\mathrm{dv}}{v}$

Note that $\int \frac{\mathrm{dv}}{v} = \ln \left(\left\mid v \right\mid\right) + C$.

$A = - \frac{1}{3} \ln \left(\left\mid v \right\mid\right) + C = - \frac{1}{3} \ln \left(\left\mid 1 + \cos \left(u\right) \right\mid\right) + C$

$A = - \frac{1}{3} \ln \left(\left\mid 1 + \cos \left(3 x\right) \right\mid\right) + C$

Now, onto the second integral:

Let $B = \int {x}^{3} {e}^{1 - {x}^{4}} \mathrm{dx}$.

Again, we will use substitution: let $t = 1 - {x}^{4}$, implying that $\mathrm{dt} = - 4 {x}^{3} \mathrm{dx}$.

$B = - \frac{1}{4} \int - 4 {x}^{3} {e}^{1 - {x}^{4}} \mathrm{dx} = - \frac{1}{4} \int {e}^{t} \mathrm{dt}$

Note that $\int {e}^{t} \mathrm{dt} = {e}^{t} + C$.

$B = - \frac{1}{4} {e}^{t} + C = - \frac{1}{4} {e}^{1 - {x}^{4}} + C$

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

$- \frac{1}{3} \ln \left(\left\mid 1 + \cos \left(3 x\right) \right\mid\right) - \frac{1}{4} {e}^{1 - {x}^{4}} + C$