Mastering Tricky Integrals: Solving sqrt(cot[x])dx with a Substitution

[julian92]

a bit tricky integral!

Homework Statement

Integral of (sqrt(cot[x])dx)

Homework Equations

I just need a hint :)

The Attempt at a Solution

 

[Mark44]

I believe that an ordinary substitution will work: u = cot(x). At least, that's where I would start.

An identity that might come in handy is cot²(x) + 1 = csc²(x).

 

[julian92]

I believe that an ordinary substitution will work: u = cot(x). At least, that's where I would start.

An identity that might come in handy is cot²(x) + 1 = csc²(x).

I think that won't work ,, bcoz i'll get Integral of (Sqrt(u)/(1+u^2))du

then what the next step?!

i also tried integration by parts after the step u mentioned ,, but it just gets more complicated

More Help Is Appreciated :)

 

[Mark44]

I think that I would go with a trig substitution next, tan w = u/1.

I haven't worked this through, so caveat emptor!

 

[physicsman2]

what I would do is convert cot x to cos x/sin x. Let u = sin x and go from there.

 

[The Chaz]

what I would do is convert cot x to cos x/sin x. Let u = sin x and go from there.

So... √(du/u) ?

 

[physicsman2]

So... √(du/u) ?

That won't work out, will it? The square root poses a problem then.

 

[julian92]

I think that I would go with a trig substitution next, tan w = u/1.

I haven't worked this through, so caveat emptor!

Umm ,,, i tried what you said ,, i think that will turn the integral back to itself :S

except i'll get Integral of Sqrt(tan(x))dx

I'm really stuck now

Thanks by the way
I appreciate it :D

and if you get any ideas i'll be glad to try some :)

 

[Cyosis]

This is quite a cumbersome integral. Start out with the substitution given by Mark in post #2 then make a substitution $y=\sqrt{u}$. Enjoy the partial fractions.

 

[Mark44]

Here's another tack: Let u = sqrt(cot(x)), du = -csc^2(x)dx/2sqrt(cot(x))

This turns the original integral into -2 int{u^2/(u^4 + 1) * du}

I think that one is amenable to a trig substitution, tan w = u^2

This is a quite messy integral!

 

[julian92]

This is quite a cumbersome integral. Start out with the substitution given by Mark in post #2 then make a substitution $y=\sqrt{u}$. Enjoy the partial fractions.

are you sure about the partial fractions? ,, because i didn't enjoy any !

i got the Integral of (2(y^2) / (1 + y^4))dy !

 

[julian92]

Here's another tack: Let u = sqrt(cot(x)), du = -csc^2(x)dx/2sqrt(cot(x))

This turns the original integral into -2 int{u^2/(u^4 + 1) * du}

I think that one is amenable to a trig substitution, tan w = u^2

This is a quite messy integral!

it tried it ,, it would also give me : (constant) times int{Sqrt(tan w) dw}

 

[Cyosis]

At this step you need to integrate a rational function. To do that use partial fractions.

 

[julian92]

At this step you need to integrate a rational function. To do that use partial fractions.

at what step exactly?!

and how can i do that ?! would you please show me how :)

 

[julian92]

hey guys ,, i managed to make the integral look like this [2][int{1/(1+(u^4)) du}]

would that be any good !

 

[Cyosis]

You need to use partial fractions to continue from:

$$-2\int \frac{y^2}{y^4+1}$$

 

[Dickfore]

First, make a substitution:

$$u = \sqrt{\cot(x)}.$$

Then you will get an integral of a rational function. The denominator is $u^{4} + 1$. In order to factorize it, you will need all the complex fourth roots of $-1$. These are:

$$\begin{array}{rcl} e^{\iota \frac{\pi}{4}} & = & \frac{1 + \iota}{\sqrt{2}} \\ e^{\iota \frac{3 \pi}{4}} & = & \frac{-1 + \iota}{\sqrt{2}} \\ e^{\iotai \frac{5 \pi}{4}} & = & \frac{-1 - \iota}{\sqrt{2}} \\ e^{\iota \frac{7 \pi}{4}} & = & \frac{1 - \iota}{\sqrt{2}} \end{array}$$

i.e. you have 2 pairs of complex conjugate roots. Therefore:

$$u^{4} + 1 = (u^{2} + \sqrt{2}u + 1)(u^{2} - \sqrt{2}u + 1)$$

From here, you should be able to proceed in the standard fashion.

 

[julian92]

You need to use partial fractions to continue from:

$$-2\int \frac{y^2}{y^4+1}$$

But i can't factor (x^4)+1

i mean it's (x^4)+1 ,, not (x^4)-1

i don't know how to use partial fractions with that !

 

[julian92]

First, make a substitution:

$$u = \sqrt{\cot(x)}.$$

Then you will get an integral of a rational function. The denominator is $u^{4} + 1$. In order to factorize it, you will need all the complex fourth roots of $-1$. These are:

$$\begin{array}{rcl} e^{\iota \frac{\pi}{4}} & = & \frac{1 + \iota}{\sqrt{2}} \\ e^{\iota \frac{3 \pi}{4}} & = & \frac{-1 + \iota}{\sqrt{2}} \\ e^{\iotai \frac{5 \pi}{4}} & = & \frac{-1 - \iota}{\sqrt{2}} \\ e^{\iota \frac{7 \pi}{4}} & = & \frac{1 - \iota}{\sqrt{2}} \end{array}$$

i.e. you have 2 pairs of complex conjugate roots. Therefore:

$$u^{4} + 1 = (u^{2} + \sqrt{2}u + 1)(u^{2} - \sqrt{2}u + 1)$$

From here, you should be able to proceed in the standard fashion.

Ahaaaa ,, that's entirely awesome :D

thanks soooooo much ,, to all of you guys :)

now ,, i can go to rest :)