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In less trivial integration problems, using trig substitution (1.1), I keep fumbling over when and what trig identities to use.

  • One trig identity I need help proving/understanding is how/why $\left[9-9sin^2\theta\right]=\left[9cos^2\theta\right]$ (?)

Once I've establish the original variables in terms of $\theta$, was is the best way, or maybe a general way if there is one, for dealing with the numerator equation? For example, after the trig substitution, is the common next step another form of integration or is it merely simplification using identities. (etc.)

In general it seems I keep drifting into harder, longer integration, so I'd like help getting a better understanding of 'the next step' following the substitution.


(1.1) Example problems for reference: $$\int\frac{8u+4}{(u^2-u)-2}du \qquad \int\frac{6z-3}{(z^2+2z)+2}dz \qquad \int\frac{\sqrt{9-x^2}}{x^2}dx$$

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  • $\begingroup$ Those parentheses in the first two integral's denominators are completely useless. Did you mean otherwise? $\endgroup$
    – DonAntonio
    Mar 13, 2014 at 22:03
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    $\begingroup$ For the trig identity, $sin^2(\theta)+cos^2(\theta)=1$. Therefore multiply by $9$ and move the $sin^2(\theta)$ to the other side. Or are you asking for a proof of $sin^2(\theta)+cos^2(\theta)=1$? $\endgroup$
    – George1811
    Mar 13, 2014 at 22:10
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    $\begingroup$ @DonAntonio The reason the given problems are in this form is to emphasize the need to complete the square. It's trivial and could as easily be omitted, it's just a form of habit and convention I use. $\endgroup$ Mar 13, 2014 at 22:13
  • $\begingroup$ @George1811 I didn't realize how obvious that was haha. That's exactly what I was asking for, thanks. $\endgroup$ Mar 13, 2014 at 22:14

1 Answer 1

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One example quick, without substitution or stuff:

$$\int\frac{8u+4}{u^2-u-2}du=4\int\frac{2u-1}{u^2-u-2}du+\frac83\int\left(\frac1{u-2}-\frac1{u+1}\right)du=$$

$$=4\log(u^2-u-2)+\frac83\left(\log\frac{u-2}{u+1}\right)+C=\ldots\text{(continue simplifying if you will)}$$

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  • $\begingroup$ I don't understand how you got the "$+ 8/3\int\dots du$" part. Could you explain where this came from? $\endgroup$ Mar 13, 2014 at 22:16
  • $\begingroup$ @ChongWaldo, you should try to do the algebra: $$\frac1{u-2}-\frac1{u+1}=\frac3{(u-2)(u+1)}$$ The $\;8\;$ comes from $$8u+4=4(2u-1)+8$$ $\endgroup$
    – DonAntonio
    Mar 14, 2014 at 6:35

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