How to Evaluate the Integral of the Square Root of 9+4x^2?

[karush]

Evaluate $\int\sqrt{9+4{x}^{2}}\ dx$
$$
\displaystyle
x= \frac{3}{2}\tan(u)
\ \ dx= \sec^2(u)
$$

 

[kaliprasad]

Evaluate $\int\sqrt{9+4{x}^{2}}\ dx$
$$
\displaystyle
x= \frac{3}{2}\sec(x)
\ \ dx= \frac{3\sin\left({x}\right)}{2\cos^2 \left({x}\right)}
$$
Then

$\displaystyle \int\sqrt{9+4 \left( \frac{3}{2} \sec{(x) }\right)^2 } \ dx
\implies 3\int\tan\left({x}\right)\ dx
$
So far?.

$\tan(x) = \dfrac{\sin(x)}{\cos(x)}$ and $\frac{d}{dx} \cos(x) = - \ sin (x)$ you should now be able to proceed

 

[karush]

I changed this looks easier
$$
\displaystyle
x= \frac{3}{2}\tan(u)
\ \ dx= \frac{3}{2}\sec^2(u)
$$

 

[MarkFL]

I would write:

\(\displaystyle I=\int \sqrt{9+4x^2}\,dx=3\int \sqrt{\left(\frac{2}{3}x\right)^2+1}\,dx\)

Let:

\(\displaystyle \frac{2}{3}x=\sinh(u)\implies dx=\frac{3}{2}\cosh(u)\,du\)

And we now have:

\(\displaystyle I=\frac{9}{2}\int \cosh^2(u)\,du=\frac{9}{4}\int \cosh(2u)+1\,du=\frac{9}{4}\sinh(u)\cosh(u)+\frac{9}{4}u+C\)

\(\displaystyle I=\frac{1}{2}x\sqrt{4x^2+9}+\frac{9}{4}\arsinh\left(\frac{2}{3}x\right)+C\)

 

[karush]

$$\displaystyle I=\frac{1}{2}x\sqrt{4x^2+9}+\frac{9}{4}\arsinh\left(\frac{2}{3}x\right)+C$$
$$\sinh\left(x\right)=y=\frac{e^{x}-e^{-x}}{2}$$
$$2y=e^{x}-e^{-x}\ \ \implies\ \ e^{2x}-2ye^{x}-1=0 $$
?
How does $\displaystyle\frac{9}{4}\arsinh\left(\frac{2}{3}x\right) $ become $\displaystyle\frac{9\ln\left({\sqrt{4x^2+9}}\right)+2x}{4 }$

 

[Prove It]

$$\displaystyle I=\frac{1}{2}x\sqrt{4x^2+9}+\frac{9}{4}\arsinh\left(\frac{2}{3}x\right)+C$$
$$\sinh\left(x\right)=y=\frac{e^{x}-e^{-x}}{2}$$
$$2y=e^{x}-e^{-x}\ \ \implies\ \ e^{2x}-2ye^{x}-1=0 $$
?
How does $\displaystyle\frac{9}{4}\arsinh\left(\frac{2}{3}x\right) $ become $\displaystyle\frac{9\ln\left({\sqrt{4x^2+9}}\right)+2x}{4 }$

Well, in short, it doesn't...

$\displaystyle \begin{align*} y &= \frac{9}{4}\,\textrm{arsinh}\,{\left( \frac{2}{3}\,x \right) } \\ \frac{4}{9}\,y &= \textrm{arsinh}\,{ \left( \frac{2}{3}\,x \right) } \\ \sinh{ \left( \frac{4}{9}\,y \right) } &= \frac{2}{3}\,x \\ \frac{1}{2}\,\left( \mathrm{e}^{\frac{4}{9}\,y} - \mathrm{e}^{-\frac{4}{9}\,y} \right) &= \frac{2}{3}\,x \\ \mathrm{e}^{\frac{4}{9}\,y} - \mathrm{e}^{-\frac{4}{9}\,y} &= \frac{4}{3}\,x \\ \mathrm{e}^{\frac{4}{9}\,y} \,\left( \mathrm{e}^{\frac{4}{9}\,y} - \mathrm{e}^{-\frac{4}{9}\,y} \right) &= \frac{4}{3}\,x\,\mathrm{e}^{\frac{4}{9}\,y} \\ \left( \mathrm{e}^{\frac{4}{9}\,y} \right) ^2 - \frac{4}{3}\,x\,\mathrm{e}^{\frac{4}{9}\,y} -1 &= 0 \\ \left( \mathrm{e}^{\frac{4}{9}\,y} \right) ^2 - \frac{4}{3}\,x\,\mathrm{e}^{\frac{4}{9}\,y} + \left(- \frac{2}{3}\,x \right) ^2 - \left( -\frac{2}{3}\,x \right) ^2 - 1 &= 0 \\ \left( \mathrm{e}^{\frac{4}{9}\,y} - \frac{2}{3}\,x \right) ^2 - \frac{4}{9}\,x^2 - 1 &= 0 \\ \left( \mathrm{e}^{\frac{4}{9}\,y } - \frac{2}{3} \,x \right) ^2 &= \frac{4}{9}\,x^2 + 1 \\ \left( \mathrm{e}^{\frac{4}{9}\,y} - \frac{2}{3} \, x \right) ^2 &= \frac{4\,x^2 + 9}{9} \\ \mathrm{e}^{\frac{4}{9}\,y} - \frac{2}{3}\,x &= \pm \frac{\sqrt{4\,x^2 + 9}}{3} \\ \mathrm{e}^{\frac{4}{9}\,y} &= \frac{2\,x \pm \sqrt{4\,x^2 + 9}}{3} \\ \frac{4}{9} \, y &= \ln{ \left( \frac{2\,x \pm \sqrt{ 4\,x^2 + 9 }}{3} \right) } \\ y &= \frac{9\ln{ \left( \frac{2\,x \pm \sqrt{ 4 \, x^2 + 9 }}{3} \right) }}{4} \end{align*}$

and in order for this to exist, we require $\displaystyle \begin{align*} y = \frac{9\ln{\left( \frac{2\,x + \sqrt{4\,x^2 + 9}}{3} \right) }}{4} \end{align*}$.

 

[karush]

good grief

thanks heavy though,
I tried and tried then cried.

where do you do your latex??
looks like math magic output

The TI Inspire gave...
$$\int \sqrt{9+4{x}^{2 }} dx
= \frac{9\ln\left({\sqrt{4{x}^{2}+9}}+2x\right)}{4 }
-\frac{x\sqrt{4{x}^{2}+9}}{2}$$

 

[Greg]

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

$$\dfrac32\sinh(u)=x,\quad u=\sinh^{-1}\left(\dfrac{2x}{3}\right)$$

$$\dfrac32\cosh(u)\,\mathrm du=\mathrm dx$$

$$\dfrac92\int\sqrt{1+\sinh^2(u)}\cosh(u)\,\mathrm du$$

$$=\dfrac92\int\cosh^2(u)\,\mathrm du$$

$$=\dfrac92\int\dfrac{e^{2x}+e^{-2x}}{4}+\dfrac12\,\mathrm du$$

$$=\dfrac94\int\cosh(2u)+1\,\mathrm du=\dfrac{9\sinh(2u)}{8}+\dfrac{9u}{4}+C_0$$

$$=\dfrac98\cdot2\cdot\dfrac{2x}{3}\cdot\dfrac{\sqrt{9+4x^2}}{3}+\dfrac{9\sinh^{-1}\left(\dfrac{2x}{3}\right)}{4}+C_0$$

$$=\dfrac{x\sqrt{9+4x^2}}{2}+\dfrac{9\ln\left(\sqrt{\left(\dfrac{2x}{3}\right)^2+1}+\dfrac{2x}{3}\right)}{4}+C_0$$

$$=\dfrac{x\sqrt{9+4x^2}}{2}+\dfrac{9\ln\left(\dfrac{\sqrt{4x^2+9}}{3}+\dfrac{2x}{3}\right)}{4}+C_0$$

$$=\dfrac{x\sqrt{9+4x^2}}{2}+\dfrac{9\ln\left(\sqrt{4x^2+9}+2x\right)}{4}-\dfrac{9\ln(3)}{4}+C_0$$

$$\int\sqrt{9+4x^2}\,\mathrm dx=\dfrac{x\sqrt{9+4x^2}}{2}+\dfrac{9\ln\left(\sqrt{4x^2+9}+2x\right)}{4}+C$$

Can you show that $\cosh\left(\sinh^{-1}(x)\right)=\sqrt{x^2+1}$ and that $\sinh^{-1}(x)=\ln(\sqrt{x^2+1}+x)$ ?

 

[Greg]

Keep going. Recall that $e^{\ln(x)}=x$.

 

[karush]

$sin^{-1} {x} =\ln\left({\sqrt{{x}^{2}+1 }+x}\right)$
$\displaystyle \frac{e^{\ln\left({\sqrt{{x}^{2}+1 }+x}\right)}
+e^{-\ln\left({\sqrt{{x}^{2}+1 }+x}\right)}}{2}
\implies
\frac{\sqrt{{x}^{2}+1}+x+\sqrt{{x}^{2}+1}-x }{2}
\implies \sqrt{{x}^{2}+1} $

 

[Greg]

\(\displaystyle \cosh\left(\sinh^{-1}(x)\right)=\dfrac{e^{\ln(\sqrt{x^2+1}+x)}+e^{-\ln(\sqrt{x^2+1}+x)}}{2}=\dfrac{e^{\ln(\sqrt{x^2+1}+x)}+\dfrac{1}{e^{\ln(\sqrt{x^2+1}+x)}}}{2}\)

\(\displaystyle =\dfrac{\sqrt{x^2+1}+x+\dfrac{1}{\sqrt{x^2+1}+x}}{2}\)

Can you get it from there?

 

[Greg]

$sin^{-1} {x} =\ln\left({\sqrt{{x}^{2}+1 }+x}\right)$
$\displaystyle \frac{e^{\ln\left({\sqrt{{x}^{2}+1 }+x}\right)}
+e^{-\ln\left({\sqrt{{x}^{2}+1 }+x}\right)}}{2}
\implies
\frac{\sqrt{{x}^{2}+1}+x+\sqrt{{x}^{2}+1}-x }{2}
\implies \sqrt{{x}^{2}+1} $

That's it with some steps omitted. Good work (though I'm not clear exactly how you arrived at the above).

 

[karush]

I used the conjugate on the second term