Question

Evaluate the integral

Answer 1

https://i.gyazo.com/52725b1e8ae5a3291468e3f693a30622.png

Answer 2

does it help to know that \[1-\sin^2(x)=\cos^2(x)\]?

Answer 3

Not really

Answer 4

how about this?

\[\Large \frac{\sin(x)}{1-\sin^2(x)}\]
\[\Large \frac{\sin(x)}{\cos^2(x)}\]
\[\Large \frac{\sin(x)}{\cos(x)*\cos(x)}\]
\[\Large \frac{1}{\cos(x)}*\frac{\sin(x)}{\cos(x)}\]
\[\Large \sec(x)*\tan(x)\]
does that help?

Answer 5

Final answer C?

Answer 6

yes because the derivative of sec(x) is sec(x)*tan(x)

Answer 7

so you have to think backwards in a way

Answer 8

Yeah, I got it.