Evaluate the definite integral, the integral of tan cube x dx divided by sec x, rages from 0 t0 pi/3?

Answer 1

Substitute: tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x)

tan^3(x) / sec(x) = sin^3(x)/cos^2(x) = sin(x)(1 - cos^2(x))/cos*2(x)

sin(x)/cos^2(x) - sin(x)

First term integrate:
let u = cos(x) and du = -sin(x)
(-1/u^2) => 1/u = 1/cos(x)

Second term integrate:
-sin(x) => cos(x)

1/cos(x) + cos(x) evaluated from 0 to pi/3
(2 + 1/2) - (1 + 1) = 1/2