#\int ((3+4sin(x)+2cos(x))/(3+2sin(x)+cos(x))) "d"x#.
Observe that the numerator and the denominator are very similar. The numerator is almost exactly double the denominator.
Therefore, add 3 and take away 3 from the numerator in order to lead to nice cancellation.
#\int ((6+4sin(x)+2cos(x)-3)/(3+2sin(x)+cos(x))) "d"x#,
#\int ((2(3+2sin(x)+cos(x)))/(3+2sin(x)+cos(x)))-3/(3+2sin(x)+cos(x)) "d"x#,
#\int (2 -3/(3+2sin(x)+cos(x))) "d"x#,
#2x - 3 \int 1/(3+2sin(x)+cos(x)) "d"x#.
Now we just have to solve,
# \int 1/(3+2sin(x)+cos(x)) "d"x#.
Often a useful substitution for tricky trigonometric integrals is a substitution called the tangent half angle substitution or the Weierstrass substitution. It is particularly useful for dealing with rational functions of #sin(x)# and #cos(x)#, as we have here.
The substitution is the following,
#t = tan(1/2x)#.
With this substitution,
#("d"t)/("d"x) = 1/2 * sec^2(1/2x)#.
Using the identity #sec^2(theta)=1+tan^2(theta)#,
#("d"t)/("d"x) = 1/2 * (1+tan^2(1/2x))#,
#("d"t)/("d"x) = 1/2 * (1+t^2)#.
We conclude that,
#"d"x = 2/(1+t^2) "d"t#.
Consider by the half angle identity for sine,
#sin(x)=2sin(x/2)cos(x/2)#
#sin(x)=2cos^2(x/2)*(sin(x/2)/cos(x/2))#
#sin(x)=(2tan(x/2))/(sec^2(x))#
#sin(x)=2t/(1+t^2)#.
Similarly,
#cos(x) = cos^2(x/2)-sin^2(x/2)#
#cos(x) = cos^2(x/2) (1-((sin(x/2))/cos(x/2))^2)#
#cos(x) = (1-tan^2(x/2))/(sec^2(x/2))#
#cos(x) = (1-t^2)/(1+t^2)#.
Substituting, # \int 1/(3+2sin(x)+cos(x)) "d"x#, is transformed to,
#\int 1/(3+(2t)/(1+t^2)+(1-t^2)/(1+t^2)) * 2/(1+t^2) "d"t#
#2\int 1/(3(1+t^2)+2t+1-t^2) "d"t#
#2\int 1/(2t^2+2t+4)"d"t#
#\int 1/(t^2+2t+2) "d"t#.
The quadratic in #t# in the denominator is irreducible (cannot be factored without the use of complex numbers), meaning that this is an arctangent integral in #t#.
#\int1/((1+t)^2+1) "d"t = arctan(t+1) + C#
Then, reversing the substitution,
# \int 1/(3+2sin(x)+cos(x)) "d"x = arctan(tan(1/2x)+1)+C#.
We conclude that,
#\int ((3+4sin(x)+2cos(x))/(3+2sin(x)+cos(x))) "d"x = 2x - 3arctan(tan(1/2x)+1) + K#.
Incidentally, you can read more about the tangent half angle substitution here.
