antiderivative????

Question

How does the antiderivative of tanx = -lncosx???? Is there some trig function or something???

Answer 1

The antiderivative of tan(x) is Int tan(x) dx = Int sin(x)/cos(x) dx. (x). Make u = cos(x), so that du = - sin(x) dx => sin(x) dx = - du. Therefore, Int tan(x) dx = Int -du/u = - ln(u) + C, C being a constant of integration. Since u = cos(x), the antideravative of the tamn function is - ln(cos(x) + C. Actually, -lncos(x) is aone of the antiderivatives, because there are infinitely many, depnding on the constant C.

Answer 2

Antiderivative is the integral

if you know anything about integrals than your find that by doing
sinx/cosx dx

say cosx = u
du = -sinx

so int(sinx/cosx)dx = int(-du/u) = -ln|u| = -ln|cosx|

Answer 3

Anti-derivative of tan(x) is found by substituting in sin(x)/cos(x). The derivative of cos(x) is -sin(x). Therefore you have -dy/y whose antiderivative is -ln(y) or when substituting in cos(x) for y you get - ln(cos(x)). Taking the derivative of this you get -1/cos(x) * d(cos(x)) which equals sin(x)/cos(x) which equals tan(x). So the answer checks out.

Answer 4

It can be solved from scratch using substitution.

∫ (tan(x) dx)

By definition, this is equal to

∫ ( [sin(x)/cos(x)] dx )

Which is equal to

∫ ([1/cos(x)] sin(x) dx)

Let u = cos(x).
du = -sin(x) dx, so
(-1) du = sin(x) dx

Note that sin(x) dx is the tail end of our integral, so it follows that
(-1) du will be the tail end after the substitution.

∫ ( (1/u) (-1) du )

Factor the (-1) from the integral,

(-1) ∫ ( (1/u) du )

Now integrate.

(-1) ln|u| + C

But u = cos(x), so our final answer is

(-1) ln|cos(x)| + C

Answer 5

well, work it out the other way: take the derivative of -ln(cosx)

d/dx of -ln(cosx) = [-1 / (cosx)]* (-sinx)
= sinx / cosx

remember from basic trig, that sinx / cosx = tanx

Answer 6

I = ∫ tan x dx
I = ∫ (sin x / cos x).dx
I = - ∫ ((-sin x) / cos x).dx
Let y = cos x
dy = - sin x.dx
I = - ∫ dy / y
I = - ln y + c
I = - ln cos x + c