Question

How do you integrate `\int \frac { \sin x \cos x } { \cos ^ { 2} x + 3\cos x + 2} d x`?

Answer 1

`lnabs(cosx+1)-2lnabs(cosx+2)+C`

Explanation:

`I=int(sinxcosx)/(cos^2x+3cosx+2)dx`

Let `u=cosx` so `du=-sinxdx`. Then, the integral becomes simpler because we can rewrite it entirely without trig functions, noting that the opposite `-du=sinxcosx` will be used:

`I=intcosx/(cos^2x+3cosx+2)(sinxdx)=-intu/(u^2+3u+2)du`

Factor the denominator.

`I=-intu/((u+1)(u+2))du`

Find the partial fraction decomposition for `u/((u+1)(u+2))`:

`u/((u+1)(u+2))=A/(u+1)+B/(u+2)`

`u=A(u+2)+B(u+1)`

Letting `u=-1` shows that:

`-1=A(-1+2)+B(-1+1)" "=>" "A=-1`

Letting `u=-2` shows that:

`-2=A(-2+2)+B(-2+1)" "=>" "B=2`

Thus:

`u/((u+1)(u+2))=(-1)/(u+1)+2/(u+2)`

Then (remember the integral had a `-` sign floating out front):

`I=int1/(u+1)du-int2/(u+2)du`

Which are both simple integrals found through using the natural logarithm:

`I=lnabs(u+1)-2lnabs(u+2)+C`

Reversing the earlier substitution:

`I=lnabs(cosx+1)-2lnabs(cosx+2)+C`