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Reply 1
There is a standard formula.
Assuming K is a constant.
Integral of coskx = 1/k x sinkx + C
So your 2cos4x would be: 2x1/4xsin4x +C = 1/2sin4x
Reply 2
you mean

Unparseable latex formula:

$\int b\cos(ax)\;dx=b\cdot\frac{1}{a} sin(ax)+c



Somit like that?
Reply 3
thanks! :smile: gosh there's so many little rules, i tend to forget most of them!
Reply 4
Just use a simple substitution of u = 4x. hence du/dx = 4, and 1/4 du = dx

2cos(4x)dx=2cos(u)14du=2.14cos(u)du=24sin(u)+C=12sin(4x)+C \int 2cos(4x) dx = 2 \int cos(u) \frac{1}{4}du = 2.\frac{1}{4} \int cos(u) du = \frac{2}{4} sin(u) + C = \frac{1}{2}sin(4x) + C

Just using substitution removes the need to memorise endless lists of general formulae.
Reply 5
Sidhe's formula is correct, yep.

Everytime you integrate a trig function like that, make sure that you differentiate it to check your answer.
Reply 6
How would you integrate 2/sec2x2/sec^{2}x please?
Reply 7
2sec2(x)=2cos2(x)\frac{2}{sec^2(x)} = 2cos^2(x)

Use an identity to find 2cos2(x)2cos^2(x) in terms of cos(2x).
Reply 8
thanks!

also, there was a cos(kx) formula given above, what's the sin(kx) one?
Reply 9
cdeu12
thanks!

also, there was a cos(kx) formula given above, what's the sin(kx) one?


It's the same only it's cos instead of sin, and it's negative.
Reply 10
notnek
2sec2(x)=2cos2(x)\frac{2}{sec^2(x)} = 2cos^2(x)

Use an identity to find 2cos2(x)2cos^2(x) in terms of cos(2x).

which identity would that be?
Reply 11
M.A.H
which identity would that be?

cos(2x)=2cos2(x)1\displaystyle cos(2x)=2cos^2(x)-1
Reply 12
I'd just solve.

Unparseable latex formula:

$2\int\cos^2(x)\;dx



But then I'm a weirdo. :smile:
Sidhe
I'd just solve.

Unparseable latex formula:

$2\int\cos^2(x)\;dx


And how are you planning on doing that? Using a trig identity is the only (quick) method of solving this integral.
Reply 14
Hashshashin
And how are you planning on doing that? Using a trig identity is the only (quick) method of solving this integral.


Like I said I'm a weirdo and I know what that equals off the top of my head. :smile: Believe it or not it's useful to know things like that, they may not be trig identitities per se but if you know them they save time.

EDIT: actually I'm lying to be frank I already know what the derivative of the original is or at least 1/2 or d/dx of sec(x)^2 of it so in actuality, so I wouldn't of even bothered using anything, I'd of just written the answer out. After a while of doing problems you tend to get a hang of looking for simple methods that don't necessarily involve basic rules.

Once you know how to derive the trig functions and have done a truck load there's no point in grinding out an answer, unless you want a bit of practice. :smile: That's what function tables and books are for, as long as you can derive them when you have to, you don't have to mess about. Of course this is absolutely no use or interest to the op, but then that's already answered.
Reply 15
how would you integrate something like cos x sin^2 x?

I've got as far as changing it to cos x multiplied by 1/2(1-2x)... but that's as far as I've managed. What would I do next and how would I go about doing it? :s-smilie: Thanks.
Reply 16
cdeu12
how would you integrate something like cos x sin^2 x?

I've got as far as changing it to cos x multiplied by 1/2(1-2x)... but that's as far as I've managed. What would I do next and how would I go about doing it? :s-smilie: Thanks.

Recognition by noting that cosx is the derivative of sinx or use a substitution - u=sinx.
Reply 17
thanks for the speedy reply notnek. i still don't really understand the recognition method, i'd be grateful it that could be explained a little more!
Reply 18
cdeu12
how would you integrate something like cos x sin^2 x?

I've got as far as changing it to cos x multiplied by 1/2(1-2x)... but that's as far as I've managed. What would I do next and how would I go about doing it? :s-smilie: Thanks.


By parts or substitution is most likely? What have you tried exactly? I'm not sure what you've done there?

Try by parts is my advice.
Reply 19
cdeu12
thanks for the speedy reply notnek. i still don't really understand the recognition method, i'd be grateful it that could be explained a little more!

If you have something of the form:

f(x)(f(x))n\displaystyle f'(x)(f(x))^{n}

Then it's integral dx is (f(x))n+1n+1+c\displaystyle \frac{(f(x))^{n+1}}{n+1} + c

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