<s>[INST] Definite Integration with Trigonometric Substitution I'm working on a question that involves using trigonometric substitution on a definite integral that will later use u substitution but I am not sure how to go ahead with this.
$$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx$$
My first step was to use $\sqrt{a^2+x^2}$ as $x=a\tan\theta$ to get...
$$2x=3\tan\theta :x=\frac32\tan\theta$$
$$dx=\frac32\sec^2\theta$$
Substituting:
$$\int\frac{\frac32\sec^2\theta}{\frac94\tan^2\theta\sqrt{9\tan^2\theta+9}}$$
The problem here is how do I change the limit it goes to?
$$\frac43=\tan\theta$$
and
$$\frac23=\tan\theta$$
Following DR.MV's answer so far..
$$\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\sec^2\theta}{\tan^2\theta\sqrt{9\sec^2\theta}}d\theta$$
$$=\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\sec\theta}{\tan^2\theta}d\theta$$
$$=\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\cos\theta}{\sin^2\theta}d\theta$$
Now $u=\sin\theta$ so $du=\cos\theta d\theta$
$$=\frac29\int_{?}^{?}\frac{1}{u^2}du$$
This is where I am stuck now...
 [/INST] $$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx = \int_a^b\frac{\frac{2}{3}\sec^2\theta}{\frac{9}{4}\tan^2\theta\sqrt{9\tan^2\theta+9}}d\theta$$
where $a = \tan^{-1}\frac{2}{3}$ and $b = \tan^{-1}\frac{4}{3}$
$$ \int_a^b\frac{2\sec\theta}{9\tan^2\theta}d\theta = \left(-\frac{2\text{cosec}\theta}{9} \right)_a^b$$
Hint: $\text{cosec}\left(\tan^{-1}\frac{4}{3}\right) = \frac{5}{4}$, $\text{cosec}\left(\tan^{-1}\frac{2}{3}\right) = \frac{\sqrt{13}}{2}$ 
EDIT: let $\tan^{-1}\frac{4}{3}=d $
$$ \frac{4}{3} = \tan d$$
$$\frac{4}{3} = \frac{\sin d}{\cos d} $$
$$ \frac{1}{\sin d} = \text{cosec}\ d = \frac{3}{4}\sec d$$
$$ \sec^2d = \tan^2 d + 1 = 1+ \left(\frac{4}{3}\right)^2 = \frac{25}{9}$$
$$\sec d =\frac{5}{3}$$
$$ \text{cosec}\ d = \frac{3}{4}\times \frac{5}{3} =\frac{5}{4}$$
You can do the same for the second one. 
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