For integral dx/(a + b*sin(x)): Use t = tan(x/2). sin(x) = 2t/(1+t^2), dx = 2dt/(1+t^2). Integral = integral 2dt/[a(1+t^2) + 2bt] = integral 2dt/[at^2 + 2bt + a]

Complete the square in denominator and integrate:

• If a^2 > b^2: result involves arctan
• If a^2 < b^2: result involves logarithm
• If a^2 = b^2: simplifies directly

For integral dx/(a + b*cos(x)): Same Weierstrass substitution. cos(x) = (1-t^2)/(1+t^2). Integral = integral 2dt/[(a+b) + (a-b)t^2]

Result when a > b > 0: integral dx/(a + b*cos(x)) = (2/sqrt(a^2-b^2)) * arctan[sqrt((a-b)/(a+b)) * tan(x/2)] + C