What's the integral of sec(x)tan^4(x)?

The integral of sec(x)tan^4(x) is (1/3)sec(x)tan^2(x) + (1/3)ln|sec(x) + tan(x)| + C.

To solve this integral, we can use the substitution u = tan(x). Then, du/dx = sec^2(x) and dx = du/sec^2(x). Substituting these into the integral, we get:

∫sec(x)tan^4(x) dx = ∫u^4/(1+u^2) du

We can then use partial fractions to split this integral into two simpler integrals:

u^4/(1+u^2) = u^2 - u^2/(1+u^2)

The first integral can be easily solved using the power rule:

∫u^2 du = u^3/3 = tan^3(x)/3

For the second integral, we can use the substitution v = 1+u^2. Then, dv/du = 2u and du = dv/2u. Substituting these into the integral, we get:

-∫(1/(2v)) dv = -1/2 ln|v| = -1/2 ln|1+u^2|

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

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

Using the identity 1+tan^2(x) = sec^2(x), we can simplify this to:

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