what is the antiderivative of (sinx)/((cosx)^2) ?

Answer 1

Integral ( [sin(x) / cos^2(x)] dx)

To solve this, we use substitution.

Let u = cos(x).
du = -sin(x) dx. Therefore

(-1)du =sin(x) dx

Our integral then becomes

Integral ( 1/(u^2) (-1) du )

Pulling out the (-1) out of the integral,

(-1) * Integral (1/(u^2) ) du

And now, solving this integral is straightforward. We use the reverse power rule. It's worth memorizing that the integral of 1/x^2 is equal to -1/x.

(-1) (-1/u) + C

(1/u) + C

Substituting back u = cos(x), we have

1/cos(x) + C
sec(x) + C

Answer 2

Antiderivatives = Undoing differentiation A function of F is termed an antiderivative of f on an period I if F' (x) = f(x) for all applications. If F is an antiderivative of f in any period I then the most familiar antiderivative of f on I is F(x)+C the position C is any consistent.