How Can We Integrate 1/(u*sqrt(1-u^2))?

[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?

 

[franznietzsche]

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.

 

[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.

 

[franznietzsche]

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).$$cosec(x) = \frac{1}{sin(x)}$$.

 

[jatin9_99]

well your subsitution is didn't work because it is difficult to solve integration with assumption not being in the question

 

[arildno]

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:
$$-\int\frac{du}{1-u^{2}}\neq\frac{-1}{2}ln(1-u^{2})+C$$

 

[speeding electron]

Fine, yes, sorry about that, me being stupid again...but that integral was getting annoying.

 

[Smin0]

Help please

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 :(.

 

[Air]

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.

 

[Smin0]

Thanks :)!

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?