<s>[INST] Using trig substitution to evaluate $\int \frac{dt}{( t^2 + 9)^2}$ $$\int \frac{\mathrm{d}t}{( t^2 + 9)^2} = \frac {1}{81} \int \frac{\mathrm{d}t}{\left( \frac{t^2}{9} + 1\right)^2}$$
$t = 3\tan\theta\;\implies \; dt = 3 \sec^2 \theta \, \mathrm{d}\theta$
$$\frac {1}{81} \int \frac{3\sec^2\theta \mathrm{ d}\theta}{ \sec^4\theta} = \frac {1}{27} \int \frac{ \mathrm{ d}\theta}{ \sec^2\theta} = \dfrac 1{27}\int \cos^2 \theta\mathrm{ d}\theta $$
$$ =\frac 1{27}\left( \frac{1}{2} \theta + 2(\cos\theta \sin\theta)\right) + C$$
$\arctan \frac{t}{3} = \theta \;\implies$
$$\frac{1}{27}\left(\frac{1}{2} \arctan \frac{t}{3} + 2 \left(\frac{\sqrt{9 - x^2}}{3} \frac{t}{3}\right)\right) + C$$
This is a mess, and it is also the wrong answer. 
I have done it four times, where am I going wrong?
 [/INST] The integral $$ \int \cos^2 \theta \, \mathrm d\theta=\frac{\theta}{2}+\frac{1}{2} \sin \theta \cos \theta+C. $$
(You had the coefficient wrong)
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