Integral of cos2x and secx.?

Question

I'm having trouble finding the integral of cos 2x and secx. can someone please help me?

can anyone explain this a little more throughly???

Answer 1

1) Use the identity:

cos(2x) = 2sin(x)cos(x)

∫ cos(2x) dx = 2 ∫sin(x)cos(x)dx

If u = sin(x), then du = cos(x)dx, so

2 ∫sin(x)cos(x)dx = 2 ∫ u du = 2[1/2 u²] + C = u² + C = sin²(x) + C

2) My calc book says "The first person to integrate sec(x) may well have spent much time doing so", which a math book's way of saying "What follows is so tricky that we wouldn't expect you to have thought of it".

First, "prepare" sec(x) by expressing it in sines and cosines:

sec(x) = 1/cos(x) = cos(x)/(cos²(x)) = cos(x)/(1 - sin²(x))

Then, use the substitution:

u = sin(x)
du = cos(x) dx

Then

∫ sec(x) dx = ∫ 1/(1 - sin²(x)) cos(x)dx = ∫ 1/(1 - u²) du

Then, as the previous poster said, you use partial fractions:

1/(1 - u²) = A/(1 - u) + B(1 + u)

1 = A(1 + u) + B(1 - u) = (A - B)u + (A + B) = 0u + 1

Then setting coefficients equal, you get 2 equations in 2 unknowns:

A - B = 0
A + B = 1
------------
2A = 1
A = 1/2
B = A = 1/2

So now the integral breaks into two pieces:

∫ 1/(1 - u²) du = 1/2 [∫ 1/(1 + u) du + ∫ 1/(1 - u) du]

And then, you could do another substitution for the denominators to show that they're going to be natural logs...but hopefully that's obvious: ∫ du/(1 + u) = ln|1 + u|, and ∫du/(1 - u) = -ln|1 - u|.

So you get:

1/2 [∫ 1/(1 + u) du + ∫ 1/(1 - u) du] = 1/2 [ln|1 + u| - ln|1 - u|] + C

And then re-substituting sin(x) for u:

1/2 [ln|1 + sin(x)| - ln|1 - sin(x)|] + C

Or, as the previous answerer mentioned, you can play with the laws of logs a little and make it into:

ln[√(|(1 + sin(x))/(1 - sin(x))|)] + C

if you like that better.

Answer 2

cos2x can be integrated using a double-angle formula. Look it up, it's actually very straightforward.

The other is harder. Try this:

secx = 1/cosx
= cosx/cos^2(x)
= cosx/(1-sin^2(x))

That last one can be integrated by a substitution of u = sinx.

well... you do have to use partial fractions after that, but it isn't that bad... your instructor wouldn't be throwing the integral of secx at you if you hadn't already learned partial fractions. Oh, and remember, before you call it finished, that the difference of logarithms turns into a logarithm of a quotient (quotient rule of logarithms)