Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 34

int (sin (omega))/(cos^3 (omega))d omega
To solve, apply u-substitution method. So, let u be:
u = cos (omega)
The, differentiate u.
du = -sin (omega) d omega
Since the sine function present in the integrand is positive, divide both sides by -1.
(du)/(-1) = (-sin (omega) d omega)/(-1)
-du= sin (omega) d omega
Plugging them, the integral becomes:
= int (-du) / u^3
= int -u^(-3) du
To take the integral of this, apply the formula int u^(n+1)/(n+1)+C .
= -u^(-2)/(-2) + C
= u^(-2)/2+C
= 1/(2u^2)+C
And, substitute back u = cos (omega) .
= 1/(2cos^2(omega)) + C

Therefore, int (sin (omega))/(cos ^3(omega)) d omega = 1/(2cos^2(omega)) + C .