Integrating tan x

Can someone show me how to integrate tan x wrt x please? I know you have 1/cos x, which gives a natural log, but after that I can't work it out. Thanks.

Also, if you could do the same for cot x that would be great!

Thanks. :)

Let J the integral you want to calculate. Let I mean "integral of..." Then,

J = I(tanx dx)

Make this change of variable:

t = tanx -> dt = dx / cos²x

Substituting,

J = I(t dt cos²x)

Now express the cos²x as function of t:

t = tanx = sinx / cosx

t² = sin²x / cos²x = (1 - cos²x) / cos²x = (1 / cos²x ) - 1

1 / cos²x = 1 + t² -> cos²x = 1 / (1 + t²)

so that

J = I(t dt / (1 + t²)) = (1/2) ln (1 + t²) = ln (Sqrt(1 + t²))

Replacing t by tanx:

J = ln(Sqrt(1 + tan²x)) = ln(1 / cos x) = -ln(cos x)

Quote:

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Originally Posted by Thomas154321

Can someone show me how to integrate tan x wrt x please? I know you have 1/cos x, which gives a natural log, but after that I can't work it out. Thanks.

Also, if you could do the same for cot x that would be great!

Thanks. :)

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tan x = (sin x)/(cos x)

Integral of (sin x)/(cos x) dx

Let u = cos x

Then du/dx = - sin x

So dx = -1/(sin x) du

Integral becomes

Integral of (sin x)/u (-1)/(sin x) du

= Integral of (-1)/u du = -ln u + c = -ln (cos x) + c

cot x = (cos x)/(sin x)

Integral of (cos x)/(sin x) dx

Let u = sin x

Then du/dx = cos x

So dx = 1/(cos x) du

Integral becomes

Integral of (cos x)/u (1)/(cos x) du

= Integral of (1/u) du = ln u + c = ln (sin x) + c

I suggest that you use the Wolfram Integrator whenever you need a quick view of the result (just to confirm whether the calculus you performed is right or not).

http://integrals.wolfram.com/index.jsp

Quote:

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Originally Posted by Rassis

I suggest that you use the Wolfram Integrator whenever you need a quick view of the result (just to confirm whether the calculus you performed is right or not).

http://integrals.wolfram.com/index.jsp

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I knew the Wolfram site but had never stumbled upon this specific page... :thumb: