Understanding less trivial integration by trig substitution.

Question

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.

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

Answer 1

One example quick, without substitution or stuff:

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

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

Answer 2

I'd say the only example out of the 3 for which the use of trig substitution is highly recommended is the integral $\int\frac{\sqrt{9-x^2}}{x^2}dx$.

$\int\frac{\sqrt{9-x^2}}{x^2}dx$

$\sin\theta=\frac{x}{3}$

$x=3\sin\theta$

$dx=3\cos\theta\ d\theta$

$\theta=\sin^{-1}\left(\frac{x}{3}\right)$

$\cos\theta=\frac{\sqrt{9-x^2}}{3}$

$\sqrt{9-x^2}=3\cos\theta$

$=\int\frac{3\cos\theta}{9\sin^2\theta}\times3\cos\theta\ d\theta$

$=\int\cot^2\theta\ d\theta$

$=\int\left(\csc^2\theta-1\right)d\theta$

$=-\cot\theta-\theta+c$

$\cot\theta=\frac{\sqrt{9-x^2}}{x}$

$=-\frac{\sqrt{9-x^2}}{x}-\sin^{-1}\left(\frac{x}{3}\right)+c$