chatterclaw73
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the integral of:
(sin(theta))/(cos(theta)-sin(theta))
I just need to know where to begin. does this integral require a substitution or maybe a trig identity to simplify it. Thanks in advance for any pointers.
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Pangol
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(Original post by chatterclaw73)
the integral of:
(sin(theta))/(cos(theta)-sin(theta))
I just need to know where to begin. does this integral require a substitution or maybe a trig identity to simplify it. Thanks in advance for any pointers.
I have some memory of seeing this, or something very like this, in a question in the past, and there being something clever that makes it straightforward. But I have no idea what that might have been right now.

You could start by multiplying numerator and denominator by cos(theta) + sin(theta) and then use a variety of trig identities and splitting up into several fractions.
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Dancer2001
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does this help as a starting point?
The first fraction can be integrated, and the second can be rearranged using double angle formulae again.
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Pangol
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(Original post by Dancer2001)
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does this help as a starting point?
The first fraction can be integrated, and the second can be rearranged using double angle formulae again.
Yeah, that's the way I did it. Doesn't end up being particularly attractive, but it works.
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chatterclaw73
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(Original post by Pangol)
I have some memory of seeing this, or something very like this, in a question in the past, and there being something clever that makes it straightforward. But I have no idea what that might have been right now.

You could start by multiplying numerator and denominator by cos(theta) + sin(theta) and then use a variety of trig identities and splitting up into several fractions.
I have tried that approach, but it became messy. I will try again and see what happens. Thanks for the quick reply.
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dextrous63
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It's been a while, but does the substitution t=tan(theta/2) deal with this? https://www.math24.net/weierstrass-substitution/
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chatterclaw73
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(Original post by dextrous63)
It's been a while, but does the substitution t=tan(theta/2) deal with this? https://www.math24.net/weierstrass-substitution/
I have also tried this. Maybe my algebra is lacking. I will try this again also. Thanks once again.
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mnot
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(Original post by chatterclaw73)
the integral of:
(sin(theta))/(cos(theta)-sin(theta))
I just need to know where to begin. does this integral require a substitution or maybe a trig identity to simplify it. Thanks in advance for any pointers.
Multiply the fraction by (cos(theta)+sin(theta)) / (cos(theta)+sin(theta))
then
sort out terms and simplify and it will become much easier.
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Pangol
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(Original post by chatterclaw73)
I have tried that approach, but it became messy. I will try again and see what happens. Thanks for the quick reply.
A bit messy, yes, but not unmanagable. It does end up requiring a couple of integrals that are usually found in A level formula books rather than ones that everyone is expected to know.
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Sir Cumference
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Here's a nice method:

Let \displaystyle I=\int \frac{\cos \theta}{\cos \theta - \sin\theta} \ d\theta and \displaystyle J = \int \frac{\sin \theta}{\cos \theta - \sin\theta} \ d\theta

Then \displaystyle I-J = \int \frac{\cos \theta - \sin \theta}{\cos \theta - \sin\theta} \ d\theta = \theta + c

And \displaystyle I+J = \int \frac{\cos \theta + \sin \theta}{\cos \theta - \sin\theta} \ d\theta

This one can be integrated by noticing that the top is -1 x the derivative of the bottom. Then solve for I and J.
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RDKGames
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(Original post by Sir Cumference)
Here's a nice method:

Let \displaystyle I=\int \frac{\cos \theta}{\cos \theta - \sin\theta} \ d\theta and \displaystyle J = \int \frac{\sin \theta}{\cos \theta - \sin\theta} \ d\theta

Then \displaystyle I-J = \int \frac{\cos \theta - \sin \theta}{\cos \theta - \sin\theta} \ d\theta = \theta + c

And \displaystyle I+J = \int \frac{\cos \theta + \sin \theta}{\cos \theta - \sin\theta} \ d\theta

This one can be integrated by noticing that the top is -1 x the derivative of the bottom. Then solve for I and J.
Lovely bit of black mathemagic. The calculus hunters would have you burning at the stake if you were to suggest this back in the day.
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Pangol
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(Original post by Sir Cumference)
Here's a nice method:

Let \displaystyle I=\int \frac{\cos \theta}{\cos \theta - \sin\theta} \ d\theta and \displaystyle J = \int \frac{\sin \theta}{\cos \theta - \sin\theta} \ d\theta

Then \displaystyle I-J = \int \frac{\cos \theta - \sin \theta}{\cos \theta - \sin\theta} \ d\theta = \theta + c

And \displaystyle I+J = \int \frac{\cos \theta + \sin \theta}{\cos \theta - \sin\theta} \ d\theta

This one can be integrated by noticing that the top is -1 x the derivative of the bottom. Then solve for I and J.
That's the one I remember seeing! This was a STEP question not too long ago, yes?
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Sir Cumference
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(Original post by Pangol)
That's the one I remember seeing! This was a STEP question not too long ago, yes?
Yes I remember something similar in STEP, not sure it was this particular integral though.
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chatterclaw73
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(Original post by Sir Cumference)
Here's a nice method:

Let \displaystyle I=\int \frac{\cos \theta}{\cos \theta - \sin\theta} \ d\theta and \displaystyle J = \int \frac{\sin \theta}{\cos \theta - \sin\theta} \ d\theta

Then \displaystyle I-J = \int \frac{\cos \theta - \sin \theta}{\cos \theta - \sin\theta} \ d\theta = \theta + c

And \displaystyle I+J = \int \frac{\cos \theta + \sin \theta}{\cos \theta - \sin\theta} \ d\theta

This one can be integrated by noticing that the top is -1 x the derivative of the bottom. Then solve for I and J.
That is a very elegant method. Thank you so much for this.
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