Indefinite Integral - II

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  • #1
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Homework Statement


[tex]\int \left(\frac{x}{x\cos x-\sin x} \right)^2 dx[/tex]


Homework Equations





The Attempt at a Solution


Factoring out ##\cos x## from the denominator, the integral transforms to
[tex]\int \sec^2x \left(\frac{x}{x-\tan x}\right)^2dx[/tex]
Substituting ##\tan x=t##, ##\sec^2 xdx=dt##
[tex]\int \left(\frac{\arctan t}{\arctan t-t}\right)^2dt[/tex]
I honestly have no idea about how to proceed from here.

Any help is appreciated. Thanks!
 

Answers and Replies

  • #2
SteamKing
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I don't think you can factor cos x from sin x and obtain tan x. I suggest you expand the terms within the brackets and see if it can be simplified with trig identities.
 
  • #3
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Why can't I factor out cosx?
sinx/cosx=tanx
 
  • #4
SteamKing
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The problem is you have only sin x in your integral.
 
  • #5
haruspex
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Look at the derivative of x cos(x) - sin(x). This suggests a way to use integration by parts.
 
  • #6
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Look at the derivative of x cos(x) - sin(x). This suggests a way to use integration by parts.
The derivative of xcos(x)-sin(x) is -xsin(x) but I still can't figure it out. The whole thing is squared, I am completely lost.
 
  • #7
SteamKing
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You could have expanded the brackets by now and checked to see if trig identities could further simplify the integrand.
 
  • #8
haruspex
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The derivative of xcos(x)-sin(x) is -xsin(x) but I still can't figure it out. The whole thing is squared, I am completely lost.
It suggested to me writing the integrand as ##\frac{x}{\sin x} \frac{x\sin x}{\left(x\cos x-\sin x \right)^2} ##. Try integrating that by parts.
 
  • #9
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It suggested to me writing the integrand as ##\frac{x}{\sin x} \frac{x\sin x}{\left(x\cos x-\sin x \right)^2} ##. Try integrating that by parts.
Integrating by parts:
[tex]\frac{x}{\sin x} \int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx-\int \frac{\sin x-x\cos x}{\sin^2 x} \left(\int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx \right)dx[/tex]

[tex]\because \int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx=(x\cos x-\sin x)[/tex]
The integral simplifies to:
[tex]\frac{x}{\sin x}(x\cos x-\sin x)+\int \left(\frac{x \cos x-\sin x}{\sin x} \right)^2dx[/tex]
If I expand the squared term, I get an integral involving cot and cot^2, should I integrate by parts again?
 
  • #10
haruspex
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[tex]\because \int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx=(x\cos x-\sin x)[/tex]
Retry that step.
 
  • #11
SammyS
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Integrating by parts:
[tex]\frac{x}{\sin x} \int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx-\int \frac{\sin x-x\cos x}{\sin^2 x} \left(\int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx \right)dx[/tex]

[tex]\because \int \frac{x\sin x}{\left(x\cos x-\sin x \right)^2}dx=(x\cos x-\sin x)[/tex]
The integral simplifies to:
[tex]\frac{x}{\sin x}(x\cos x-\sin x)+\int \left(\frac{x \cos x-\sin x}{\sin x} \right)^2dx[/tex]
If I expand the squared term, I get an integral involving cot and cot^2, should I integrate by parts again?
[STRIKE]How is that integration by parts? What were the parts you used?[/STRIKE]

Try ##\displaystyle \ u=\frac{x}{\sin(x)}\ \text{ and }\ dv=\frac{x\sin(x)}{\left(x\cos(x)-\sin(x) \right)^2}\,dx \ . ##

Added in Edit: Part of the above edited out.
I see it now .
 
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  • #12
haruspex
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How is that integration by parts? What were the parts you used?

Try ##\displaystyle \ u=\frac{x}{\sin(x)}\ \text{ and }\ dv=\frac{x\sin(x)}{\left(x\cos(x)-\sin(x) \right)^2}\,dx \ . ##
Looks to me that Pranav-Arora was getting there - just made a mistake in the first integration.
 
  • #13
SammyS
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Looks to me that Pranav-Arora was getting there - just made a mistake in the first integration.
Oh ! I see it now.

I had looked at it several times and didn't see how that fit integration by parts.
 
  • #14
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Retry that step.
Oops, its the reciprocal of that.

This time I get:
[tex]\frac{x}{\sin x}\frac{1}{x\cos x-\sin x}+\int \csc^2 xdx=\frac{x}{\sin x}\frac{1}{x\cos x-\sin x}-\cot x[/tex]

But on further simplification, I do not end up with the right answer.
 
  • #15
ehild
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It is correct, bring to common denominator and simplify with sinx.Show what you did.

ehild
 
  • #16
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It is correct, bring to common denominator and simplify with sinx.Show what you did.

ehild
Got it, thanks! :smile:

Writing cot in terms cos and sin
[tex]\frac{x-x\cos^2 x+\cos x \sin x}{\sin x(x\cos x-\sin x)}=\frac{x(1-\cos^2 x)+\cos x \sin x}{\sin x(x\cos x-\sin x)}[/tex]
As 1-cos^2x=sin^2x
[tex]\frac{x\sin^2 x+\cos x \sin x}{\sin x(x\cos x-\sin x)}=\frac{x\sin x+\cos x }{x\cos x-\sin x}[/tex]

Thanks a lot haruspex! :smile:
 
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