I know that the intregation of [math]\frac{sinx}{cos^2x}[/math] is [math]secx[/math].

 

But I'm not sure how to get to the answer. I started off by using the trigonometric identity: [math]cos^2x + sin^2x = 1[/math].

 

Therefore, [math]\frac{sinx}{1-sin^2x}[/math]. Now what?

[math]\int \frac{sinx}{cos^2x}dx[/math]

using regular substitution, let [math]u=cosx[/math]

You mean like this: [math]\int (sinx)(u^{-2})[/math]?

If we let [math]u = \cos x[/math] then [math] du = -\sin x \, dx[/math]; hence our integral is transformed into:

 

[math]-\int \frac{1}{u^2} \, du = \sec u + c[/math]