For integral $\frac{dx}{a + b*sin(x}$): Use t = tan$\frac{x}{2}$. sin(x) = 2$\frac{t}{1+t^2}$, dx = 2$\frac{dt}{1+t^2}$. 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 $\frac{dx}{a + b*cos(x}$): Same Weierstrass substitution. cos(x) = $\frac{1-t^2}{(1+t^2)}$. Integral = integral 2dt/[(a+b) + (a-b)$t^2$]

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