Question

What is the antiderivative of `(sinx)/(cos^2x)`?

Answer 1

`\sec x + C`

Explanation:

The simplest way to do this is to recognise the patterns, namely that `(\cos^n x)' = n \cos^{n-1} x (- \sin x) = - n \cos^{n-1} x \ \sin x`.

here `n-1 = -2 \implies n = -1`

so we can explore `(\cos^{-1} x)' = (-1) \cos^{-2} x (- \sin x) = \frac{\sin x}{\cos^{2} x}` ... hey presto!

so `\int \frac{\sin x}{\cos^{2} x} dx = \frac{1}{\cos x} + C = \sec x + C`

A more methodical way, if you are just starting out, would be, starting with the integral `\int \frac{\sin x}{\cos^{2} x} dx`, to make the sub `u = \cos x, du = -\sin x dx`

So the integral becomes:

`\int \frac{\sin x}{\cos^2 x} (-\frac{1}{\sin x}) du = -\int \frac{1}{u^2} du`

`= \frac{1}{u} + C = \frac{1}{\cos x} + C = \sec x + C`