MHB Integral: I am considering a trigonometric substitution

[karush]

$\int\frac{3{x}^{3}}{\sqrt{4{x}^{2}-1}}dx $

I wasn't sure what substitution to use due to what is in radical?

$x=\frac{1}{2}\sec^2 \left({\theta}\right)\ dx=\frac{\sin\left({\theta}\right)}{\cos^3\left({}\right)}$

 

[MarkFL]

I think I would consider IBP here...

$$I=\int\frac{3x^3}{\sqrt{4x^2-1}}\,dx=\frac{3}{8}\int x^2\cdot\frac{8x}{\sqrt{4x^2-1}}\,dx$$

Can you proceed?

 

[karush]

Shouldn't
$x=\frac{1}{2} \sec\left({\theta}\right)$

 

[MarkFL]

Shouldn't
$x=\frac{1}{2} \sec\left({\theta}\right)$

You could try using that substitution, but you will wind up with something as difficult if not more so (if my calculations are correct), than the original. I really think IBP is the way to go here. :D

 

[karush]

But then how do we get rid of the radical?

To get $tan x$

 

[MarkFL]

This is what I have in mind...

$$I=\int\frac{3x^3}{\sqrt{4x^2-1}}\,dx=\frac{3}{8}\int x^2\cdot\frac{8x}{\sqrt{4x^2-1}}\,dx$$

Now, using IBP, we let:

$$u=x^2\implies du=2x\,dx$$

$$dv=\frac{8x}{\sqrt{4x^2-1}}\,dx\implies v=2\sqrt{4x^2-1}$$

And so we now have:

$$I=\frac{3}{8}\left(2x^2\sqrt{4x^2-1}-\frac{1}{2}\int \sqrt{4x^2-1}8x\,dx\right)$$

Now integrating the remaining integral is just a matter of applying the power rule. :D

 

[karush]

Well that's pretty cool
Thank you

However thot by parts could be avoided
Guess not

I'll proceed to finish

 

[Dethrone]

If you want to use a trigonometric substitution, then you will need to evaluate $3/64 \int \sec^4 \theta \,d\theta$ which can be easily evaluated by splitting $\sec^4 \theta=\sec^2 \theta(1+\tan^2\theta)$.

 

[Prove It]

$\int\frac{3{x}^{3}}{\sqrt{4{x}^{2}-1}}dx $

I wasn't sure what substitution to use due to what is in radical?

$x=\frac{1}{2}\sec^2 \left({\theta}\right)\ dx=\frac{\sin\left({\theta}\right)}{\cos^3\left({}\right)}$

I think Mark is right that Integration by Parts will be the most direct method. I might also be inclined to try a HYPERBOLIC substitution $\displaystyle \begin{align*} x = \frac{1}{2}\cosh{(t)} \implies \mathrm{d}x = \frac{1}{2}\sinh{(t)}\,\mathrm{d}t \end{align*}$, giving

$\displaystyle \begin{align*} \int{ \frac{3x^3}{\sqrt{4x^2 - 1}}\,\mathrm{d}x} &= \int{ \frac{3 \left[ \frac{1}{2}\cosh{(t)} \right] ^3 }{\sqrt{ 4 \left[ \frac{1}{2}\cosh{(t)} \right] ^2 - 1 }} \,\frac{1}{2}\sinh{(t)}\,\mathrm{d}t } \\ &= \int{ \frac{ \frac{3}{8}\cosh^3{(t)} }{ \sqrt{ \cosh^2{(t)} - 1 } } \, \frac{1}{2}\sinh{(t)}\,\mathrm{d}t } \\ &= \frac{3}{16} \int{ \frac{\cosh^3{(t)} \sinh{(t)}}{\sinh{(t)}}\,\mathrm{d}t } \\ &= \frac{3}{16} \int{ \cosh^3{(t)}\,\mathrm{d}t } \\ &= \frac{3}{16} \int{ \left[ 1 + \sinh^2{(t)} \right] \, \cosh{(t)}\,\mathrm{d}t } \end{align*}$

and this can be finished with a very simple substitution $\displaystyle \begin{align*} u = \sinh{(t)} \implies \mathrm{d}u = \cosh{(t)}\,\mathrm{d}t \end{align*}$. See how you go.

 

[karush]

By u substitution then

$\frac{3}{16}\int\left[1-{{u}^{2 }}\right]du
= \frac{3}{16}\left[u-\frac{{u}^{3}}{3}\right]+C$

 

[Prove It]

By u substitution then

$\frac{3}{16}\int\left[1-{{u}^{2 }}\right]du
= \frac{3}{16}\left[u-\frac{{u}^{3}}{3}\right]+C$

Yes so now convert back to a function of x...

 

[karush]

$u=\sinh\left({t}\right)$
$$x=\frac{1}{2}\cosh\left({t}\right)$$

Plugging back in ?

 

[Prove It]

$u=\sinh\left({t}\right)$
$$x=\frac{1}{2}\cosh\left({t}\right)$$

Plugging back in ?

Surely you can at least plug in u in terms of t. Then use the identity $\displaystyle \begin{align*} \cosh^2{(t)} - \sinh^2{(t)} \equiv 1 \end{align*}$ so that you can write $\displaystyle \begin{align*} \sinh{(t)} = \sqrt{\cosh^2{(t)} - 1} \end{align*}$. That will be enough to write your answer in terms of x...

 

[karush]

$\displaystyle \begin{align*} u=\sinh{(t)}
= \sqrt{\cosh^2{(t)} - 1} \end{align*}$

$$\frac{3}{16}\int\left[1-{{u}^{2 }}\right]du
=\frac{3}{16}\left[u-\frac{{u}^{3}}{3}\right]+C
=\frac{3}{16}\left[\sqrt{\cosh^2{(t)} - 1}
-\frac{{(\cosh^2{(t)} - 1})^{3/2}}{3}\right]+C
$$

this looks weird?

 

[Prove It]

$\displaystyle \begin{align*} u=\sinh{(t)}
= \sqrt{\cosh^2{(t)} - 1} \end{align*}$

$$\frac{3}{16}\int\left[1-{{u}^{2 }}\right]du
=\frac{3}{16}\left[u-\frac{{u}^{3}}{3}\right]+C
=\frac{3}{16}\left[\sqrt{\cosh^2{(t)} - 1}
-\frac{{(\cosh^2{(t)} - 1})^{3/2}}{3}\right]+C
$$

this looks weird?

It looks weird because you are not finished! Surely you can see that since $\displaystyle \begin{align*} x = \frac{1}{2}\cosh{(t)} \end{align*}$ then $\displaystyle \begin{align*} \cosh{(t)} = 2x \end{align*}$...

 

[karush]

$$\frac{3}{16}\left[\sqrt{4x^2 - 1}
-\frac{{(4x^2 - 1})^{3/2}}{3}\right]+C$$

ok the TI gave

$$\int\frac{3{x}^{3}}{\sqrt{4{x}^{2}-1}}dx
=\frac{\left(2{x}^{2}+1\right)\sqrt{4{x}^{2}-1}}{8}+C$$

so ?

 

[Prove It]

$$\frac{3}{16}\left[\sqrt{4x^2 - 1}
-\frac{{(4x^2 - 1})^{3/2}}{3}\right]+C$$

ok the TI gave

$$\int\frac{3{x}^{3}}{\sqrt{4{x}^{2}-1}}dx
=\frac{\left(2{x}^{2}+1\right)\sqrt{4{x}^{2}-1}}{8}+C$$

so ?

Have you tried anything to see if the two expressions are in fact equivalent?

 

[karush]

They might be =
But I couldn't get an identity.