`int sin(omega)/(cos^3(omega)) d omega` Evaluate the integral

`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` .