int costheta/(3+2sintheta+sin^2theta) d theta Use integration tables to find the indefinite integral.

Indefinite integral are written in the form of int f(x) dx = F(x) +C
 where: f(x) as the integrand
           F(x) as the anti-derivative function 
           C  as the arbitrary constant known as constant of integration
The evaluate the given integral problem: int cos(theta)/(3+2sin(theta)+sin^2(theta)) d theta , we may apply u-substitution by letting: u =sin(theta) then du = cos(x) dx .
Plug-in u =sin(theta) then du = cos(x) dx , the integral becomes:
int cos(theta)/(3+2sin(theta)+sin^2(theta)) d theta =int (cos(theta) d theta)/(3+2sin(theta)+sin^2(theta))
=int (du)/(3+2u+u^2) orint (du)/(u^2+2u+3)
It resembles a formula from table of integrals:
int 1/(ax^2+bx+c) dx = 2/sqrt(4ac-b^2)arctan((2ax+b)/sqrt(4ac-b^2)) +C
By comparing ax^2+bx+c with u^2+2u+3 , we have: a=1 , b =2and c=3 .
Plug-in the values on the integral formula, we get:
int (du)/(u^2+2u+3) =2/sqrt(4(1)(3)-(2)^2)arctan((2(1)u+2)/sqrt(4(1)(3)-(2)^2)) +C
=2/sqrt(12-4)arctan((2u+2)/sqrt(12-4)) +C
=2/sqrt(8)arctan((2u+2)/sqrt(8)) +C
=2/(2sqrt(2))arctan(2(u+1)/(2sqrt(2))) +C
= 1/sqrt(2)arctan((u+1)/sqrt(2))+C
Plug-in  u =sin(theta) on 1/sqrt(2)arctan((u+1)/sqrt(2))+C , we get the indefinite integral as:
int cos(theta)/(3+2sin(theta)+sin^2(theta)) d theta= 1/sqrt(2)arctan((sin(theta)+1)/sqrt(2))+C