1 1 Cosx 1 1 Cosx

1 1 Cosx 1 1 Cosx

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Integrate (ane/(one-cos x)) dx


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    ae4jm

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

Homework Statement

[tex]\int \frac{1}{one-cos x} dx[/tex]

Homework Equations

The Endeavour at a Solution

I found in the book where they used the [tex]cos x= \frac{1-z^2}{1+z^2}[/tex], but we haven’t went over this in class however. I’m wandering what my other options are to integrate this trouble? Can anyone delight give me a good start on finding this integral? Thanks for your aid!

Answers and Replies

A pocket-sized hint: make the bottom equal to (sinx)^2. Yous’ll have to then employ reciprocal trig functions after that.

Integrate (one/1-cosx))dx

Thanks for your help.

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

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

The possibilities and so far are making cos 10=ane/sec x, 1=cos x/cox x, and I don’t think that (sin x)^2 is =1/(csc x)^ii

What did I do wrong? Practise I demand to substitute?

You don’t need to make a commutation. Think nearly how you can brand that integral into an integral containing nothing but reciprocal trigonometric functions.

[tex]\int \frac{1}{ane-cos x} dx[/tex]

Hi ae4jm!
:smile:

Simplest method is to rewrite it as:

[tex]\int \frac{1}{2.sin^2(10/2)} dx\,,[/tex]​

which you lot should instantly recognise equally the derivative of … ?
:smile:

… standard trig equation …

This is a standard trig equation which you should know:

[tex]cos2x\,=\,cos^2x – sin^2x[/tex]

[tex]=\,2cos^2x\,-\,1[/tex]

[tex]=\,1\,-\,2sin^2x[/tex]​

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

Multiply numerator and denominator by:

[tex]1+\cos x[/tex]

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 accept:

integral (1+cos x)/(i-cos^2 x) dx and that is when I tried using trig identities and I am nevertheless just non seeing the simplification steps.

LOL, I think that possibly my light isn’t turning on this night.

[tex]\int\frac{1}{1-\cos 10}dx[/tex]

[tex]\int\frac{i+\cos x}{1-\cos^two 10}dx[/tex]

[tex]\int\left(\frac{1}{\sin^2 x}+\frac{\cos ten}{\sin^2 10}\right)dx[/tex]

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

Terminal edited:

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

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

LOL, I apparently demand to review my derivatives :p Only what’s important is that you can do it either way 🙂

Nifty! A big thank you 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 so I plugged the trouble that I began with into the calculator and told it to discover the integral from .25 to .5 and information technology’due south answer with it’s integral was the same every bit my answer with my integral.

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So, does -cscx-cotx look proficient to you lot guys?

Cheers for all of the help, I really needed information technology on this one!

To check if your reply is right, have the derivative of the solution to your anti-derivative and run into if it matches your original Integral.

Note that

[tex]-\cot \frac{x}{two} = -\sqrt{\frac{1 + \cos x}{1 – \cos x}} = -\sqrt{\frac{(1 + \cos x)^2}{1 – \cos^two x}} = -\frac{1 + \cos 10}{\sin x} = -\csc x – \cot x[/tex]

where inspection shows nosotros demand the positive square root, then the ii are equivalent. That Integrator is one astonishing piece of software.

Hey ae4jm!
:smile:

Have y’all tried these all the same?

[tex]cos2x\,=\,cos^2x – sin^2x[/tex]

[tex]=\,2cos^2x\,-\,1[/tex]

[tex]=\,ane\,-\,2sin^2x[/tex]​

(And also, of class: sin2x = two.cox.sinx)
:smile:

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

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

[tex]-\cot \frac{x}{2} = -\sqrt{\frac{1 + \cos 10}{1 – \cos ten}} = -\sqrt{\frac{(1 + \cos x)^2}{i – \cos^2 x}} = -\frac{1 + \cos x}{\sin 10} = -\csc ten – \cot x[/tex]

[tex]cot(x/2)\,=\,2cos^2(x/2)/2sin^2(ten/ii)\,=\,(i\,-\,cosx)/sinx\,=\,cosecx\,+\,cotx\,;[/tex]​

and
yous don’t have any ambiguous square-roots,

so y’all never need to say:

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

Now

try using those formula on the original integral:
:smile:

[tex]\int \frac{1}{1-cos x} dx\,.[/tex]​

plzz dummies !!! the simplest and preices solution is

first take Y=tan(x/ii)

then differentiate it to get dx=(ii/one+y^two)dy

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

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

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

oh remember ane thing besides sinx = 2y/(ane+y^2)………..its the mathematical proof values u can consult with ur teachersssss likewise

i hope this volition aid u …..thanks byeee….

plzz dummies !!! the simplest and preices solution is

first take Y=tan(x/2)

then differentiate it to go dx=(ii/ane+y^2)dy

then take cos ten = (1-y^ii)/(i+y^2)

then put higher up values in equation of dx/(1+cosx) .and so solve it by yourself its enough !!! ohh ane thingi have forgotten ….. the answer volition be (tan(10/2)+c)

similarly u tin likewise evaluate dx/(1+cosx)………..

oh call back ane thing besides sinx = 2y/(1+y^2)………..its the mathematical proof values u can consult with ur teachersssss as well

i hope this volition assist u …..thank you byeee….

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