How can I integrate (1/(1-cos x)) dx without using cos x= (1-z^2)/(1+z^2)?

[ae4jm]

[SOLVED] Integrate (1/(1-cos x)) dx

Homework Statement

\int \frac{1}{1-cos x} dx

Homework Equations

The Attempt at a Solution

I found in the book where they used the cos x= \frac{1-z^2}{1+z^2}, but we haven't went over this in class yet. I'm wandering what my other options are to integrate this problem? Can anyone please give me a good start on finding this integral? Thanks for your help!

 

[Snazzy]

A small hint: make the bottom equal to (sinx)^2. You'll have to then use reciprocal trig functions after that.

 

[ae4jm]

Integrate (1/1-cosx))dx

Thanks for your help.

Now I've got the integral of (1+cosx)/(sin x)^2 dx

Will I use a u substitution after applying the reciprocal identities.

The possibilities so far are making cos x=1/sec x, 1=cos x/cox x, and I don't think that (sin x)^2 is =1/(csc x)^2

What did I do wrong? Do I need to substitute?

 

[Snazzy]

You don't need to make a substitution. Think about how you can make that integral into an integral containing nothing but reciprocal trigonometric functions.

 

[tiny-tim]

\int \frac{1}{1-cos x} dx

Hi ae4jm!

Simplest method is to rewrite it as:

\int \frac{1}{2.sin^2(x/2)} dx\,,​

which you should instantly recognise as the derivative of … ?

 

[ae4jm]

How did you get it to this problem? I know that there will know be the multiplied 1/2 in front of the integral.

 

[tiny-tim]

… standard trig equation …

This is a standard trig equation which you should know:

cos2x\,=\,cos^2x - sin^2x

=\,2cos^2x\,-\,1

=\,1\,-\,2sin^2x​

(And also, of course: sin2x = 2.cox.sinx)

 

[rocomath]

Multiply numerator and denominator by:

1+\cos x

 

[ae4jm]

I started with trying to find the integral of 1/(1-cosx) dx.

I then multiplied the numerator and denominator by 1+cosx and I then have:

integral (1+cos x)/(1-cos^2 x) dx and that is when I tried using trig identities and I am still just not seeing the simplification steps.

LOL, I think that maybe my light isn't turning on tonight.

 

[rocomath]

\int\frac{1}{1-\cos x}dx

\int\frac{1+\cos x}{1-\cos^2 x}dx

\int\left(\frac{1}{\sin^2 x}+\frac{\cos x}{\sin^2 x}\right)dx

Continue simplifying ... use a Trig identity for the left and the right simply needs a u-sub.

 

[Snazzy]

Doesn't necessarily need a sub if you can remember what the integrals of (cscx)^2 and cotxcscx are.

 

[rocomath]

Doesn't necessarily need a sub if you can remember what the integrals of (cscx)^2 and cotxcscx are.

LOL, I obviously need to review my derivatives :p But what's important is that you can do it either way :)

 

[ae4jm]

Great! A big thanks to Rocomath, Snazzy, and Tiny-Tim!

I got -cscx-cotx. The integrator from Wolfram got -cot(x/2).

So, I plugged my -cscx-cotx into my TI-84 and integrated from .25 to .5 and got 4.042 and then I plugged the problem that I began with into the calculator and told it to find the integral from .25 to .5 and it's answer with it's integral was the same as my answer with my integral.

So, does -cscx-cotx look good to you guys?

Thanks for all of the help, I really needed it on this one!

 

[rocomath]

To check if your answer is correct, take the derivative of the solution to your anti-derivative and see if it matches your original Integral.

 

[Tedjn]

Note that

-\cot \frac{x}{2} = -\sqrt{\frac{1 + \cos x}{1 - \cos x}} = -\sqrt{\frac{(1 + \cos x)^2}{1 - \cos^2 x}} = -\frac{1 + \cos x}{\sin x} = -\csc x - \cot x

where inspection shows we need the positive square root, so the two are equivalent. That Integrator is one amazing piece of software.

 

[tiny-tim]

Hey ae4jm!

Have you tried these yet?

cos2x\,=\,cos^2x - sin^2x

=\,2cos^2x\,-\,1

=\,1\,-\,2sin^2x​

(And also, of course: sin2x = 2.cox.sinx)

(Obviously, you need to do them with x instead of 2x.)

For example, using them, you get a simpler version of:

-\cot \frac{x}{2} = -\sqrt{\frac{1 + \cos x}{1 - \cos x}} = -\sqrt{\frac{(1 + \cos x)^2}{1 - \cos^2 x}} = -\frac{1 + \cos x}{\sin x} = -\csc x - \cot x

cot(x/2)\,=\,2cos^2(x/2)/2sin^2(x/2)\,=\,(1\,-\,cosx)/sinx\,=\,cosecx\,+\,cotx\,;
​

and you don't have any ambiguous square-roots, so you never need to say:

where inspection shows we need the positive square root, so the two are equivalent.

Now try using those formula on the original integral:

\int \frac{1}{1-cos x} dx\,.
​

 

[muhammadali12]

pleasez dummies ! the simplest and preices solution is

first take Y=tan(x/2)

then differentiate it to get dx=(2/1+y^2)dy

and then take cos x = (1-y^2)/(1+y^2)

then put above values in equation of dx/(1+cosx) .then solve it by yourself its enough ! ohh one thingi have forgotten ... the answer will be (tan(x/2)+c)


similarly u can also evaluate dx/(1+cosx)...

oh remember one thing as well sinx = 2y/(1+y^2)...its the mathematical proof values u can consult with ur teachersssss as well

i hope this will help u ...thanks byeee...

 

[muhammadali12]

pleasez dummies ! the simplest and preices solution is

first take Y=tan(x/2)

then differentiate it to get dx=(2/1+y^2)dy

and then take cos x = (1-y^2)/(1+y^2)

then put above values in equation of dx/(1+cosx) .then solve it by yourself its enough ! ohh one thingi have forgotten ... the answer will be (tan(x/2)+c)


similarly u can also evaluate dx/(1+cosx)...

oh remember one thing as well sinx = 2y/(1+y^2)...its the mathematical proof values u can consult with ur teachersssss as well

i hope this will help u ...thanks byeee...