MHB Indefinite integral using trig substitutions

[karush]

$\int\frac{1}{\sqrt{2+3y^2}}dy$
$u=\sqrt{3/2}\tan\left({\theta}\right)$

I continued but it went south..

 

[Euge]

Hi karush,

First write $\sqrt{2 + 3y^2}$ as $\sqrt{3}\sqrt{\frac{2}{3} + y^2}$ in the integrand, and let $y = \sqrt{2/3}\tan \theta$. See what you get from this substitution.

 

[karush]

$\sqrt{3}\sqrt{2/3+2/3 \tan^2\left({\theta}\right)}$

 

[Euge]

Re: int using trig subs

Now use the identity $1+tan^2(\theta) = \sec^2(\theta)$ to simplify. What is $dy$ in terms of $d\theta$?

 

[karush]

$du=\frac{\sqrt{6}}{2}\sec^2\left({\theta}\right)d\theta$

 

[Euge]

Re: int using trig subs

You're close but it's $dy = \frac{\sqrt{6}}{3}\sec^2(\theta)\, d\theta$. Now express your integral in terms of the $\theta$ variable.

 

[Prove It]

Even though the trig sub will work, the integral will still be quite difficult. $\displaystyle \begin{align*} y = \sqrt{\frac{2}{3}}\sinh{(t)} \end{align*}$ is a much easier sub...

 

[Euge]

Even though the trig sub will work, the integral will still be quite difficult. $\displaystyle \begin{align*} y = \sqrt{\frac{2}{3}}\sinh{(t)} \end{align*}$ is a much easier sub...

Perhaps, but I guess it depends on whether karush knows hyperbolic functions yet. I suspect that at this stage, he already knows $\int \sec x \, dx = \ln|\sec x + \tan x| + C$, so he doesn't have to work through any difficult steps.

 

[karush]

My confusion seems to be getting $dy=dx$
Initially I factored out 2 to get $(1+\frac{3}{2}y^2)$

 

[Euge]

Wait, you're going backwards. Let's continue. We have $y = \sqrt{2/3}\tan(\theta)$, so $\sqrt{2+3y^2} = \sqrt{2}\sec(\theta)$ and $dy = \sqrt{2/3}\sec^2(\theta) \, d\theta$. Thus

$$\int \frac{1}{\sqrt{2+3y^2}}\, dy = \int \frac{1}{\sqrt{2}\sec(\theta)}\, (\sqrt{2/3} \sec^2(\theta)\, d\theta) = \frac{1}{\sqrt{3}} \int \sec(\theta)\, d\theta = \frac{1}{\sqrt{3}} \ln|\sec(\theta) + \tan(\theta)| + C.$$

Now revert back to the $y$-variable.

 

[karush]

$\frac{1}{\sqrt{3}}\ln\left({\sqrt{2+3y^2}+\sqrt{3}y}\right)+C$

 

[Prove It]

$\int\frac{1}{\sqrt{2+3y^2}}dy$
$u=\sqrt{3/2}\tan\left({\theta}\right)$

I continued but it went south..

Using the substitution I suggested:

$\displaystyle \begin{align*} \int{ \frac{1}{\sqrt{2 + 3y^2}}\,\mathrm{d}y} &= \frac{1}{\sqrt{3}}\int{ \frac{1}{\sqrt{ \frac{2}{3} + y^2}}\,\mathrm{d}y} \end{align*}$

Now substitute $\displaystyle \begin{align*} y = \sqrt{ \frac{2}{3} } \sinh{(t)} \implies \mathrm{d}y = \sqrt{\frac{2}{3}} \cosh{(t)}\,\mathrm{d}t \end{align*}$ and the integral becomes

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

Now remembering that $\displaystyle \begin{align*} y = \sqrt{ \frac{2}{3} }\sinh{(t)} \end{align*}$ that means

$\displaystyle \begin{align*} y &= \frac{\sqrt{2}\sinh{(t)}}{\sqrt{3}} \\ \sqrt{3}\,y &= \sqrt{2}\sinh{(t)} \\ \frac{\sqrt{3}\,y}{\sqrt{2}} &= \sinh{(t)} \\ \frac{\sqrt{6}\,y}{2} &= \sinh{(t)} \\ t &= \textrm{arsinh}\,\left( \frac{\sqrt{6}\,y}{2} \right) \end{align*}$

and thus the answer is $\displaystyle \begin{align*} \frac{\sqrt{3}}{3} \,\textrm{arsinh}\,\left( \frac{\sqrt{6}\,y}{2} \right) + C \end{align*}$.

 

[Euge]

$\frac{1}{\sqrt{3}}\ln\left({\sqrt{2+3y^2}+\sqrt{3}y}\right)+C$

Good work, karush! That's completely correct.