1. ## integration by substitution

$\int\frac{1}{\sqrt{9+4x^2}}dx ($Let $x=\frac{3}{2}tan\theta)$

$\frac{dx}{d\theta}=\frac{3}{2}sec^2\theta$

$\int\frac{1}{\sqrt{9+4x^2}}dx=\int\frac{1}{\sqrt{9 +4(\frac{3}{2}tan^2\theta)}} d\theta$

$=\frac{3}{18}\int\frac{1+tan^2\theta}{1+tan^2\thet a}d\theta$

$=\frac{3}{18}\theta+c$

2. ## Re: integration by substitution

Your substitution is ok. Now calculate:
$\sqrt{9+4x^2}=\sqrt{9+4\left[\frac{3}{2}\tan(\theta)\right]^2}$
$=\sqrt{9+9\tan^2(\theta)}=3\cdot \sqrt{1+\tan^2(\theta)}=\frac{3}{\cos(\theta)}$

The integral becomes:
$\int \frac{\frac{3}{2\cos^2(\theta)}}{\frac{3}{\cos(\th eta)}}d(\theta)$
$=\int \frac{\cos(\theta)}{2\cos^2(\theta)}d(\theta)}= \frac{1}{2} \int \frac{d(\theta)}{\cos(\theta)}$

Now you can use the t-formulas or the substitution, let $\theta=\arcsin(t)$.

3. ## Re: integration by substitution

it will be easier to use hyperbolic substitution
let x = (3/2) sinh (theta)
gives dx = (3/2)cosh (theta) d(theta)

4. ## Re: integration by substitution

Originally Posted by Punch
$\int\frac{1}{\sqrt{9+4x^2}}dx ($Let $x=\frac{3}{2}tan\theta)$

$\frac{dx}{d\theta}=\frac{3}{2}sec^2\theta$

$\int\frac{1}{\sqrt{9+4x^2}}dx=\int\frac{1}{\sqrt{9 +4(\frac{3}{2}tan^2\theta)}} d\theta$

you forgot to square 3/2 and substitute for dx
$\int\frac{1}{\sqrt{9+4x^2}}dx =$

$\int\frac{\frac{3}{2}\sec^2{\theta}}{\sqrt{9+4 \left(\frac{9}{4}\tan^2{\theta} \right)}} \, d\theta =$

$\int \frac{3}{2} \cdot \frac{\sec^2{\theta}}{3\sqrt{1 + \tan^2{\theta}}} \, d\theta =$

$\frac{1}{2} \int \sec{\theta} \, d\theta =$

$\frac{1}{2} \ln|\sec{\theta}+\tan{\theta}| + C$

since $x = \frac{3}{2} \tan{\theta}$ ...

$\tan{\theta} = \frac{2x}{3}$

$\sec{\theta} = \sqrt{1 + \frac{4x^2}{9}}$

back substitute ...

$\frac{1}{2} \ln \left|\sqrt{1 + \frac{4x^2}{9}} + \frac{2x}{3} \right| + C$

5. ## Re: integration by substitution

Hello, Punch!

$\int\frac{dx}{\sqrt{9+4x^2}}$

$\text{Let }2x \,=\, 3\tan\theta \quad\Rightarrow\quad dx \,=\,\tfrac{3}{2}\sec^2\!\theta\,d\theta$

. . $\sqrt{9+4x^2} \:=\:\sqrt{9 + 9\tan^2\theta} \:=\;\sqrt{9(1 + \tan^2\!\theta)} \:=\:\sqrt{9\sec^2\!\theta} \:=\:3\sec\theta$

Substitute: . $\int \frac{\frac{3}{2}\sec^2\!\theta\,d\theta}{3\sec \theta} \;=\; \tfrac{1}{2}\int\sec\theta\,d\theta \;=\;\tfrac{1}{2}\ln|\sec\theta + \tan\theta| + C$

Back-substitute: . $\tan\theta \,=\,\frac{2x}{3},\;\;\sec\theta \,=\,\frac{\sqrt{9+4x^2}}{3}$

We have: . $\frac{1}{2}\ln\left|\frac{\sqrt{9+4x^2}}{3} + \frac{2x}{3}\right| + C \;=\;\boxed{\frac{1}{2}\ln\left|\frac{2x + \sqrt{9+4x^2}}{3}\right| + C}$

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This can be simplified even further . . .

$\tfrac{1}{2}\ln\left|\frac{2x + \sqrt{9+4x^2}}{3}\right| + C$

. . $=\;\tfrac{1}{2}\bigg[\ln\left|2x+\sqrt{9+4x^2}\right| - \ln 3\bigg] + C$

. . $=\;\tfrac{1}{2}\ln\left|2x + \sqrt{9+4x^2}\right|\, \underbrace{-\, \tfrac{1}{2}\ln 3 + C}_{\text{a constant!}}$

. . $=\;\tfrac{1}{2}\ln\left|2x + \sqrt{9+4x^2}\right| + C$