Evaluate the integral of sec^3(x) dx.

The integral of sec^3(x) dx is (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C.

To evaluate the integral of sec^3(x) dx, we can use integration by substitution. Let u = sec(x) + tan(x), then du/dx = sec(x)tan(x) + sec^2(x). Rearranging this equation, we get sec^2(x) = du/dx - sec(x)tan(x). Substituting this into the integral, we get:

∫sec^3(x) dx = ∫sec(x)sec^2(x) dx
= ∫sec(x)(du/dx - sec(x)tan(x)) dx
= ∫du - u tan(x) dx
= u - ln|sec(x) + tan(x)| + C
= sec(x) + tan(x) - ln|sec(x) + tan(x)| + C

However, we want the integral in terms of sec(x) only. To do this, we can use the identity sec^2(x) - 1 = tan^2(x), which gives us:

u = sec(x) + tan(x)
u^2 - 1 = sec^2(x) + 2sec(x)tan(x) + tan^2(x) - 1
= sec^2(x) + tan^2(x) + 2sec(x)tan(x)
= sec^2(x)(1 + tan^2(x)) + 2sec(x)tan(x)
= sec^2(x)sec^2(x) + 2sec(x)tan(x)
= sec^4(x) + 2sec(x)tan(x)

Substituting this into the integral, we get:

∫sec^3(x) dx = (1/2)∫(sec^4(x) + 2sec^2(x)tan(x)) dx
= (1/2)(1/3)sec^3(x) + (1/2)ln|sec(x) + tan(x)| + C
= (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C

Therefore, the integral of sec^3(x) dx is (1/2)sec(x)tan(x) + (1