How to integrate 1/(1+sin(x))?

To integrate 1/(1+sin(x)), use the substitution u = tan(x/2).

To integrate 1/(1+sin(x)), we can use the substitution u = tan(x/2). This substitution is useful because it allows us to express sin(x) in terms of u. We have:

sin(x) = 2tan(x/2)/(1+tan^2(x/2)) = 2u/(1+u^2)

We can also express dx in terms of u:

dx = 2dx/(1+tan^2(x/2)) = 2du/(1+u^2)

Substituting these expressions into the integral, we get:

∫1/(1+sin(x)) dx = ∫(1+u^2)/(3+2u^2) du

We can now use partial fractions to simplify this expression:

(1+u^2)/(3+2u^2) = A + Bu

Multiplying both sides by 3+2u^2 and equating coefficients, we get:

A = 1/3, B = -1/3

Substituting these values back into the integral, we get:

∫1/(1+sin(x)) dx = ∫(1/3 - u/3) du = (1/3)ln|3+2u^2| - (1/6)ln|1+u^2| + C

Substituting back u = tan(x/2), we get the final answer:

∫1/(1+sin(x)) dx = (1/3)ln|3+2tan^2(x/2)| - (1/6)ln|1+tan^2(x/2)| + C