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Forgot how to Integrate cosec x

ssadi

I forgot how to integrate cosec x dx. Any help please. :embarrassed:

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Reply 1

ssadi

I forgot how to integrate cosec x dx. Any help please. :embarrassed:

...

Reply 2

charlie123

will its 1/sinx so use the chain rule,

Reply 3

ssadi

charlie123

will its 1/sinx so use the chain rule,

I am trying to do that, and failing. Please say what I should differentiate here to be able to integrate 1/sinx

Reply 4

Totally Tom

ach, morons in this forum is increasing exponentially.

use substitution

t=\tan{\frac{x}{2}}

which gives

\sin{x}=\frac{2t}{1+t^2}

Reply 5

Morbo

ssadi

I forgot how to integrate cosec x dx. Any help please. :embarrassed:

Let

y = \cos x

This produces:

\int \mathrm{cosec}x dx = -\int \frac{1}{1-y^2} dy

,

which can be solved by partial fractions, giving:

= \frac{1}{2} \mathrm{ln}\left(\frac{1-\cos x}{1+\cos x}\right) + C

.

This can be rearranged using

\mathrm{sec}^2\theta - \mathrm{tan}^2\theta = 1

to give the answer as

\mathrm{ln}|\mathrm{tan}\left(\frac{x}{2}\right)| + C

Reply 6

ssadi

Morbo

Let

y = \cos x

This produces:

\int \mathrm{cosec}x = -\int \frac{1}{1-y^2} dy

,

which can be solved by partial fractions, giving:

= \frac{1}{2} \mathrm{ln}\left(\frac{1-\cos x}{1+\cos x}\right) + C

.

This can be rearranged using

\mathrm{sec}^2\theta - \mathrm{tan}^2\theta = 1

to give the answer as

\mathrm{ln}|\mathrm{tan}\left(\frac{x}{2}\right)| + C

cosec x then equals

\frac{1}{\sqrt(1-y^2)}

how do i solve it by partial fractions then?

Reply 7

ssadi

Totally Tom

ach, morons in this forum is increasing exponentially.

use substitution

t=\tan{\frac{x}{2}}

which gives

\sin{x}=\frac{2t}{1+t^2}

I have never used the method for any other sum, is it ok for fp1?

Reply 8

generalebriety

ssadi

I have never used the method for any other sum, is it ok for fp1?

Yes. t = tan(x/2) works quite well for a lot of trig integration.

Reply 9

ssadi

generalebriety

Yes. t = tan(x/2) works quite well for a lot of trig integration.

Good to know.

Reply 10

ssadi

generalebriety

Yes. t = tan(x/2) works quite well for a lot of trig integration.

What is your opinion of the other method that the alien shows us in the thread?

Reply 11

hippobluecampe

ssadi

What is your opinion of the other method that the alien shows us in the thread?

HAHAHAHAHAHA.
Sorry that is pretty funny.

Reply 12

ssadi

Syafiquaaa

HAHAHAHAHAHA.
Sorry that is pretty funny.

I forgot his name while quoting, and had to resort what I remembered, thanks to his signature.

Reply 13

generalebriety

ssadi

What is your opinion of the other method that the alien shows us in the thread?

*shrugs* It looks fine to me.

The integral of cosec x, if it's not in the formula book, is one you should expend the effort remembering anyway.

Reply 14

Morbo

ssadi

cosec x then equals

\frac{1}{\sqrt(1-y^2)}

how do i solve it by partial fractions then?

Don't forget to change the variable of integration i.e. find dx in terms of dy. That gives you the integral I wrote.

Then partial fractions just makes

\frac{1}{1-y^2} \equiv \frac{1}{(1+y)(1-y)} \equiv \frac{A}{1+y} + \frac{B}{1-y}

. Identify

A = \frac{1}{2}

and

B = \frac{1}{2}

and integrate as usual.

Reply 15

ssadi

Morbo

Don't forget to change the variable of integration i.e. find dx in terms of dy. That gives you the integral I wrote.

Then partial fractions just makes

\frac{1}{1-y^2} \equiv \frac{1}{(1+y)(1-y)} \equiv \frac{A}{1+y} + \frac{B}{1-y}

. Identify

A = \frac{1}{2}

and

B = \frac{1}{2}

and integrate as usual.

How do you remove the square root?

Reply 16

Morbo

ssadi

How do you remove the square root?

\displaystyle \int \mathrm{cosec}x dx = \int \frac{1}{\mathrm{sin}x} dx = \int \frac{1}{\sqrt{1-y^2}} \frac{dy}{\sqrt{1-y^2}} = \int \frac{1}{1-y^2} dy

Reply 17

Zygroth

Totally Tom

t=\tan{\frac{x}{2}}

which gives

\sin{x}=\frac{2t}{1+t^2}

How do you get that?!

Reply 18

Morbo

Zygroth

How do you get that?!

\displaystyle \\ \frac{2t}{1+t^2} = \frac{2 \tan\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)}[br]\\ = \frac{2 \tan\left(\frac{x}{2}\right)}{\mathrm{sec}^2\left(\frac{x}{2}\right)}[br]\\ = 2\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)[br]\\ = \sin x

Reply 19

ssadi

Morbo

\displaystyle \int \mathrm{cosec}x dx = \int \frac{1}{\mathrm{sin}x} dx = \int \frac{1}{\sqrt{1-y^2}} \frac{dy}{\sqrt{1-y^2}} = \int \frac{1}{1-y^2} dy

I have never before converted dx into dy. I do not still get how you convert it, as I do not see any dy around that can be used to swap places :confused: