How to integrate sec(x)tan^3(x)?

To integrate sec(x)tan^3(x), use substitution u = tan(x) and convert the integral to a rational function.

Let u = tan(x), then du/dx = sec^2(x) dx and dx = du/sec^2(x). Substituting these into the integral, we get:

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

Using partial fractions, we can write the integrand as:

u^3/(1 + u^2) = u - u/(1 + u^2)

Therefore, the integral becomes:

∫ sec(x)tan^3(x) dx = ∫ (u - u/(1 + u^2)) du/sec^2(x)

Integrating the first term, we get:

∫ u/sec^2(x) du = tan(x) + C1

For the second term, we can use the substitution v = 1 + u^2, then dv/du = 2u and du/dv = 1/2(1 + u^2)^(-1/2). Substituting these, we get:

∫ -u/(1 + u^2) sec^2(x) dx = -1/2 ∫ du/(1 + u^2)^(-1/2) = -1/2 arcsin(u) + C2

Substituting back u = tan(x), we get:

∫ sec(x)tan^3(x) dx = tan(x) - 1/2 arcsin(tan(x)) + C

Therefore, the solution is:

∫ sec(x)tan^3(x) dx = tan(x) - 1/2 arcsin(tan(x)) + C, where C = C1 + C2.