Simple Intregation

[EvoN1020v]

I know that the intregation of $\frac{sinx}{cos^2x}$ is $secx$.

 

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

 

Therefore, $\frac{sinx}{1-sin^2x}$. Now what?

[NeonBlack]

$\int \frac{sinx}{cos^2x}dx$

using regular substitution, let $u=cosx$

[EvoN1020v]

You mean like this: $\int (sinx)(u^{-2})$?

[Dave]

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

 

$-\int \frac{1}{u^2} \, du = \sec u + c$