How Is the Integral of sqrt(y^2-16)/y Solved Using Trigonometric Substitution?

[karush]

206.8.4.31
Given (answer by maxima)
$$\displaystyle
I_{31}=\int \frac{\sqrt{y^2-16}}{y} \, dy =
4\arcsin\left(\dfrac{4}{\left|y\right|}\right)+\sqrt{y^2-16}+C
$$
trig subst
$$y=4\tan\left({\theta}\right)
\therefore dy=4\sec^2 (\theta)d\theta$$
so
$$\displaystyle I_{31}=4\int\frac{\sec^3(\theta)}{\tan\left({\theta}\right)}\,d\theta$$

kinda ?

 

[MarkFL]

It seems you are implying that:

$$\sec^2(\alpha)=\tan^2(\alpha)-1$$

and that's not an identity. :D

 

[karush]

206.8.4.31
Given (answer by maxima)
$$\displaystyle
I_{31}=\int \frac{\sqrt{y^2-16}}{y} \, dy =
4\arcsin\left(\dfrac{4}{\left|y\right|}\right)+\sqrt{y^2-16}+C
$$
trig substitution use identity $\tan^2\theta = \sec^2\theta - 1 $
$$y=4\sec\left({\theta}\right)
\therefore dy=4\sin\theta\sec^2 (\theta) \, d\theta \\
\displaystyle\theta=\sec^{-1}\left(\frac{y}{4}\right)$$
so
$$\displaystyle
I_{31}=\int\frac{\sqrt{16 \sec^2{\theta}-16}}{4\sec\theta}4\sin\theta\sec^2\theta \,d\theta
=4\int\tan^2\left({\theta}\right) \, d\theta$$

so far ?

 

[MarkFL]

Looks good so far...:D

 

[karush]

206.8.4.31
Given (answer by maxima)
$$\displaystyle
I_{31}=\int \frac{\sqrt{y^2-16}}{y} \, dy =
4\arcsin\left(\dfrac{4}{\left|y\right|}\right)+\sqrt{y^2-16}+C
$$
identity
$\tan^2\theta = \sec^2\theta - 1 $
$$y=4\sec\left({\theta}\right)
\therefore dy=4\sin\theta\sec^2 (\theta) \, d\theta \\
\displaystyle\theta=\sec^{-1}\left(\frac{y}{4}\right)$$
so
$$\displaystyle
I_{31}=\int\frac{\sqrt{16 \sec^2{\theta}-16}}{4\sec\theta}4\sin\theta\sec^2\theta \,d\theta
=4\int\tan^2\left({\theta}\right) \, d\theta$$
$$=\frac{-4
\left[\theta \cos\left({\theta}\right)-\sin\left({\theta}\right)
\right]}
{\cos\left({\theta}\right)}
=-4\theta +\tan\left({\theta}\right)+C $$
back subst $\theta=
\displaystyle\theta=\sec^{-1}\left(\frac{y}{4}\right)$

$$I_{31}=-4\arccos\left[\frac{4}{y}\right]+\sqrt{y^2-16}+C$$

Got a sign and arcsin vs arccos ?

 

[MarkFL]

Hint: Consider the following identity:

$$\arccos(u)+\arcsin(u)=\frac{\pi}{2}$$

 

[karush]

$\tiny{206.8.4.31} \\
\text{Given (answer by maxima)}$
$$\displaystyle
I_{31}=\int \frac{\sqrt{y^2-16}}{y} \, dy =
4\arcsin\left(\dfrac{4}{\left|y\right|}\right)+\sqrt{y^2-16}+C
$$
$\text{identity }
\displaystyle\tan^2\theta = \sec^2\theta - 1 $
$$\displaystyle y=4\sec\left({\theta}\right)
\therefore dy=4\sin\theta\sec^2 (\theta) \, d\theta \\
\displaystyle\theta=\sec^{-1}\left(\frac{y}{4}\right)$$
$\text{substitute}$
$$\displaystyle
I_{31}=\int\frac{\sqrt{16 \sec^2{\theta}-16}}{4\sec\theta}4\sin\theta\sec^2\theta \,d\theta
=4\int\tan^2\left({\theta}\right) \, d\theta$$
$$=\frac{-4
\left[\theta \cos\left({\theta}\right)-\sin\left({\theta}\right)
\right]}
{\cos\left({\theta}\right)}
=-4\theta +\tan\left({\theta}\right)+C $$
$\text{back substitute }
\displaystyle \theta=\sec^{-1}\left(\frac{y}{4}\right)$
$$I_{31}=4\arcsin \left[\frac{4}{y}\right]+\sqrt{y^2-16}+C$$

 

[karush]

Hint: Consider the following identity:

$$\arccos(u)+\arcsin(u)=\frac{\pi}{2}$$

That helped.

 

[Prove It]

206.8.4.31
Given (answer by maxima)
$$\displaystyle
I_{31}=\int \frac{\sqrt{y^2-16}}{y} \, dy =
4\arcsin\left(\dfrac{4}{\left|y\right|}\right)+\sqrt{y^2-16}+C
$$
trig subst
$$y=4\tan\left({\theta}\right)
\therefore dy=4\sec^2 (\theta)d\theta$$
so
$$\displaystyle I_{31}=4\int\frac{\sec^3(\theta)}{\tan\left({\theta}\right)}\,d\theta$$

kinda ?

I would have made some manipulation to the integrand first to make the substitutions a little more obvious...

$\displaystyle \begin{align*} \int{ \frac{\sqrt{y^2 - 16}}{y}\,\mathrm{d}y} &= \int{ \frac{\left( \sqrt{y^2 - 16} \right) ^2}{y\,\sqrt{y^2 - 16}}\,\mathrm{d}y } \\ &= \int{ \frac{\left( \sqrt{y^2 - 16} \right) ^2 }{y^2} \left( \frac{y}{\sqrt{y^2 - 16}} \right) \,\mathrm{d}y } \end{align*}$

Now let $\displaystyle \begin{align*} u = \sqrt{y^2 - 16} \implies \mathrm{d}u = \frac{y}{\sqrt{y^2 - 16}}\,\mathrm{d}y \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{\left( \sqrt{y^2 - 16} \right) ^2}{y^2}\left( \frac{y}{\sqrt{y^2 - 16}} \right)\,\mathrm{d}y } &= \int{ \frac{u^2}{u^2 + 16}\,\mathrm{d}u } \\ &= \int{ \left( 1 - \frac{16}{u^2 + 16} \right) \,\mathrm{d}u } \\ &= \int{1\,\mathrm{d}u} - \int{ \frac{16}{u^2 + 16}\,\mathrm{d}u } \end{align*}$

and I'm sure you can continue from here...