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How do you find ∫ cscx dx?

- Thread starter PrudensOptimus
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How do you find ∫ cscx dx?

Science Advisor

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Look it up in an integral table? {ln(tan(x/2))}

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(Your textbook almost certainly shows how to compute the integral of

∫ csc x dx = ∫ csc x * (csc x + cot x)/(csc x + cot x) dx

=∫ ((csc x)^2 + csc x cot x)/(csc x + cot x) dx

=-ln | csc x + cot x | + constant

hope it answers your question....

=∫ ((csc x)^2 + csc x cot x)/(csc x + cot x) dx

=-ln | csc x + cot x | + constant

hope it answers your question....

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Here's how I would do it: csc(x)= 1/sin(x) so this is the same as ∫ dx/sin(x)- an odd power of x. Back in calculus, I learned that when you have an odd power of either sin(x) or cos(x), you can use one of them with dx and use sin

∫sin(x)dx/sin

Now let u= cos(x) so du= sin(x)dx and the integeral becomes

∫ du/(1-u

The integral is now (1/2)∫du/(1-u)+ (1/2)∫du/(1+u)= -(1/2)ln|1-u|+ (1/2)ln|1+u|+C= (1/2)ln(|1+u|/|1-u|)+ C= (1/2)ln((1+sin(x))/(x-sin(x)))+ C= ln(√((1+sin(x))/(1-sin(x)))+ C

Hmmm, maybe Reihhard-t's solution isn't so scary after all!

but the most important ,when you do math,you must try the most simple,short, and clear....and get the best result.

what I did actually just multiply with 1 which is ∫(csc x + cot x)/(csc x + cot x) ....and then just carry two more steps,and done...

that's math all about

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Here's what I had to far, I had 2 approaches

∫ 1/sinx dx = ∫ (1/sinx)*cosx/cosx dx

= ∫ (1/u)(1/cosx) du = ∫1/sqrt(u^2 - u^4) du.

I don't know how to continue from there, or is it even continuable at all?

2nd approach:

∫(1/sinx) dx = ∫sinx/sin^2x dx = ∫sinx/(1-cos^2x) dx

assuming u = cos x, du = -sinx

= ∫ 1/(u^2-1) du

Then I stop there... don't know how to continue. Or is it continuable??

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You can use partial fractions to continue that, once you learn them.

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HallsofIvy lost a negative u= cos(x) so du= -sin(x)dx

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Originally posted by Hurkyl

You can use partial fractions to continue that, once you learn them.

Can you teach me? Or someone please teach me?

so I recommend you to implement my previos method.that one is actually a math trick.it is fast and incisive

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Integral of

1/sinx = sinx/(1-cos^2x) = 1/(u^2-1) du = 2/(u-1) - 2/(u+1)

So the integral is 2*(ln(u-1) - ln(u+1)) + C

Is that correct?

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Also, don't forget your absolute value signs!

And, finally, you need to back substitute to change the 'u's back into 'x's.

it becomes (1/2)ln |(cos x - 1)/(cos x + 1)|

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