# Integrating cosec...

by speeding electron
Tags: cosec, integrating
 P: 65 Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=-du/sqrt(1-u^2) Int(cosec x)dx = -Int{1/sqrt(1-u^2)}^2}du = -Int {du/(1-u^2)} = -(1/2)ln(1-u^2) + C = -(1/2)ln(sin^[2]x) +C = ln(cosec x) +C Yet differentiating back gives -cot x. Why does this substitution not work?
P: 1,783
 Quote by speeding electron Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=du/sqrt(1-u^2) Int(cosec x)dx = Int{1/sqrt(1-u^2)}^2}du = Int {du/(1-u^2)} =(1/2)ln(1-u^2) + C =(1/2)ln(sin^[2]x) +C =ln(sin x) +C Yet differentiating back gives cot x. Why does this substitution not work?

First:

$$csc x = \frac{1}{sin x}$$

NOT cos x. So your substitution is wrong to begin with. If you correct that the integration should go like this:

$$u = sin x dx$$

$$du = cos x dx$$

$$sec x du = dx$$

$$\frac{1}{\sqrt{1-u^2}} du = dx$$

So this makes the integral:

$$\int \frac{1}{u\sqrt{1-u^2}} du$$

See if you can take it from there.
 P: 65 My query was concerning the substitution u = cos x , rather than u = sin x . I did make a mistake, arcos x = -1/sqrt(1-x^2). I've edited my original post.
P: 1,783
Integrating cosec...

 Quote by speeding electron My query was concerning the substitution u = cos x , rather than u = sin x . I did make a mistake, arcos x = -1/sqrt(1-x^2). I've edited my original post.

That is the wrong substitution for integrating cosec(X).$$cosec(x) = \frac{1}{sin(x)}$$.
 P: 15 well your subsitution is didn't work because it is difficult to solve integration with assumption not being in the question
HW Helper
PF Gold
P: 12,016
 Quote by speeding electron Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=-du/sqrt(1-u^2) Int(cosec x)dx = -Int{1/sqrt(1-u^2)}^2}du = -Int {du/(1-u^2)} = -(1/2)ln(1-u^2) + C = -(1/2)ln(sin^[2]x) +C = ln(cosec x) +C Yet differentiating back gives -cot x. Why does this substitution not work?
Because you make an equality out of the following non-equality:
$$-\int\frac{du}{1-u^{2}}\neq\frac{-1}{2}ln(1-u^{2})+C$$
 P: 65 Fine, yes, sorry about that, me being stupid again...but that integral was getting annoying.
P: 8
 Quote by franznietzsche First: $$csc x = \frac{1}{sin x}$$ NOT cos x. So your substitution is wrong to begin with. If you correct that the integration should go like this: $$u = sin x dx$$ $$du = cos x dx$$ $$sec x du = dx$$ $$\frac{1}{\sqrt{1-u^2}} du = dx$$ So this makes the integral: $$\int \frac{1}{u\sqrt{1-u^2}} du$$ See if you can take it from there.
I can't :(. I've spent most of this afternoon trying this question and some other one. I know I could just use the set result, but I want to understand it, and I don't see how to integrate that last result at all :(.
 P: 206 There is a special method to this. $\int \mathrm{cosec} x \ \mathrm{d}x$ If you multiply this by $\frac {\mathrm{cosec} x - \cot x}{\mathrm{cosec} x - \cot x}$ and simplify the numerator you will get an integral of... $\int \frac{\mathrm{cosec} ^2 x - \mathrm{cosec} x \cot x}{\mathrm{cosec} x - \cot x} \ \mathrm{d}x$ Substitute $u = \mathrm{cosec} x - \cot x$ and it should work out beautifully. Carry on from here and post back if you still need help.
 P: 8 Thanks, that worked out really well :). I was wondering though, how would one work it out from the form $$\int \frac{1}{u\sqrt{1-u^2}} du$$ ? Could anyone help me see how to integrate it from this?

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