Trigonometric Integration: Understanding the Magic Behind Integration Techniques

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mateomy
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This is more of a trig question but its coming out of an integration problem I am dealing with...


the initial problem is...

[tex] \int\frac{\sqrt(1+x^2)}{x}dx[/tex]

I've worked through the problem with the book and down the line I have to integrate csc(theta) without using a handy-dandy table of integrals. Anyway, the book shows the following...

[tex] \int csc\theta \frac{csc\theta - cot\theta}{csc\theta - cot\theta} d\theta[/tex]

I can't justify to myself why they did that, let alone what magical hat they pulled that from. Can someone show me the light? Thanks.
 
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mateomy said:
This is more of a trig question but its coming out of an integration problem I am dealing with...


the initial problem is...

[tex] \int\frac{\sqrt(1+x^2)}{x}dx[/tex]

I've worked through the problem with the book and down the line I have to integrate csc(theta) without using a handy-dandy table of integrals. Anyway, the book shows the following...

[tex] \int csc\theta \frac{csc\theta - cot\theta}{csc\theta - cot\theta} d\theta[/tex]

I can't justify to myself why they did that, let alone what magical hat they pulled that from. Can someone show me the light? Thanks.
What they've done is to multiply the integrand by 1, which you can always do, and which doesn't change the value of the integrand.


The reason they did this is that the numerator is [csc2(theta) - csc(theta)cot(theta)] d(theta), which just happens to be the differential of the denominator, which is -cot(theta) + csc(theta).

You can make the substitution u = -cot(theta) + csc(theta), so du = csc2(theta) - csc(theta)cot(theta). This leads to an easy integral.
 
Okay, I can see it. I also understand what you're saying by multiplying by 1. My problem is just HOW they got that particular equivalent of 1? Was it just being clever? That's not an identity right?

Thanks, btw.
 
I mean...just the simple {csc + cot}, not the whole rational.
 
Oh okay, (sigh of relief)...I thought my mind was gone.