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...
 A: There was a typo in the current post.  After enforcing the substitution $2x=3\tan \theta$, the integral ought to read
$$\begin{align}I&=\int_{\arctan(2/3)}^{\arctan(4/3)}\frac{\frac32 \sec^2\theta}{\frac94 \tan^2\theta\sqrt{9\tan^2\theta+9}}d\theta\\\\
&=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\frac{ \sec^2\theta}{ \tan^2\theta\,\sec \theta}d\theta\\\\
&=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\frac{ \sec^2\theta}{ \tan^2\theta\,\sec \theta}d\theta\\\\
&=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\cot \theta \csc \theta d\theta\\\\
&=\frac29 \left.(-\csc \theta)\right|_{\arctan(2/3)}^{\arctan(4/3)}\\\\
&=\frac{\sqrt{13}}{9}-\frac{5}{18}\end{align}$$

NOTES:
Remark 1:  When making a substitution of variables in a definite integral, the limits of integration change accordingly.  In this example, the substitution was $x=\frac32 \tan \theta$.  When $x=1$ at the lower limit, $\tan \theta =\frac23\implies \theta =\arctan(2/3)$.  Similarly,  when $x=2$ at the upper limit, $\tan \theta =\frac43\implies \theta =\arctan(4/3)$.
Remark 2:
To evaluate $\sin (\arctan(2/3))$, we recall that the arctangent is an angle whose tangent is $2/3$.  A picture sometimes facilitates the analysis wherein we draw a right triangle with vertical side of length $2$ and horizontal side of length $3$ forming a right angle. 
Note that the angle the hypotenuse makes with the horizontal side is $\arctan(2/3)$.  Inasmcuh as the hypotenuse is of length $\sqrt{2^2+3^2}=\sqrt{13}$, we see $\sin(\arctan(2/3))=\frac{2}{\sqrt{13}}$ and thus $\csc (\arctan(2/3))=\frac{\sqrt{13}}{2}$.
A: \begin{align*}\int\frac{\frac32\sec^2\theta\,\mathrm d\mkern1.5mu\theta}{\frac94\tan^2\theta\sqrt{9\tan^2\theta+9}}&=\frac29\int\frac{\mathrm d\mkern1.5mu\theta}{\sin^2\theta\sqrt{1+\tan^2\theta}}=\frac29\int\frac{\lvert\cos\theta\rvert\,\mathrm d\mkern1.5mu\theta}{\sin^2\theta}\\[1ex]
&=\frac29\int\frac{\cos\theta\,\mathrm d\mkern1.5mu\theta}{\sin^2\theta}\qquad\text{since}\enspace 0\le\theta<\dfrac\pi2\\[1ex]
&=-\frac2{9\sin\theta}
\end{align*}
Some trigonometry will let you determine the bounds for $\sin\theta\;$ from the bounds for $\tan\theta$: since $0\le\theta<\dfrac\pi2$, we have:
$\cos\theta=\dfrac1{\sqrt{1+\tan^2\theta}}$, hence $$\sin\theta=\tan\theta\cos\theta=\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}=\frac{\cfrac{2x}3}{\sqrt{1+\cfrac{4x^2}9}}=\frac{2x}{\sqrt{4x^2+9}}$$
so that the indefinite integral is:
$$-\frac{\sqrt{4x^2+9}}{9x}.$$
A: $$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx = \int_a^b\frac{\frac{2}{3}\sec^2\theta}{\frac{9}{4}\tan^2\theta\sqrt{9\tan^2\theta+9}}d\theta$$
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. 
