I want to solve the below:
$$int frac{sqrt{16x^2 – 9}}{x} , dx$$
I know for trig substitution, if I have something in the form of $sqrt{x^2a^2}$, I can use $x = asec{u}$; it just so happens my integral has a numerator in this form: $sqrt{16x^2 – 3^2}$ so I know to use $x = 3sec u$:
$$
begin{align}
& int frac{sqrt{16x^2 – 9}}{x} , dx \
= {} & int frac{sqrt{16x^2 – 3^2}}{x} , dx \
= {} & int frac{sqrt{16(3sec u)^2 – 3^2}}{3sec u} 3sec utan u , du \
= {} & int frac{(sqrt{16(3sec u)^2 – 3^2)}(3sec utan u)}{3sec u} , du \
= {} & int sqrt{(16(3sec u)^2 – 3^2)}(tan u) , du
end{align}
$$
This doesn’t seem to make it easy. However, using a calculator online, it suggests I instead use $x = dfrac{3}{4}sec{u}$ which simplifies the integral to a crisp $int 3tan^2 u , du$.
My question is, how did the calculator get $a = dfrac{3}{4}$ and is there a way to determine an ideal trig substitution for a given function?
In general, if you have $sqrt{p x^2pm q}$
 Make the coefficient $x$ equal to $1$ by taking coefficient of $x^2$ out of square root which gives $$sqrt{px^2pm q}=sqrt psqrt{x^2pm frac{q}{p}}$$

Above expression: $sqrt{x^2pm frac{q}{p}}$ can be changed into the form: $sqrt{x^2pm a^2}$ by equating $a=sqrt{dfrac{q}{p}}$

Substitute $x=asec u$ for the form $sqrt{x^2a^2}$ and $x=atan u$ for the form $sqrt{x^2+a^2}$
For this case: $$sqrt{16x^29}=sqrt{16}sqrt{x^2frac{9}{16}}$$ $$sqrt{x^2a^2}=sqrt{x^2frac{9}{16}}$$
$$implies a=sqrt{frac{9}{16}}=frac34$$
Note: $sqrt{16x^23^2}$ is a difference of squares. Draw a picture of a right triangle suggested by this: $4x$ the hypotenuse, $3$ one of the legs (say the side opposite angle $theta$), and $sqrt{16x^23^2}$ the side adjacent to angle $theta$.
Do it, don’t just rely on my description.
So then:
$$
sintheta = frac{3}{4x},
\
costheta = frac{sqrt{16x^23^2}}{4x},
\
tantheta = frac{3}{sqrt{16x^23^2}}.
$$
Use the simplest one to suggest the substitution:
$$
x = frac{3}{4}csc theta,
\
dx = frac{3}{4}cscthetacottheta;dtheta
$$
Then substitute back into your integral, looking at your picture to find how to move between $x$ and $theta$. Here
$$
frac{sqrt{16x^2 – 9}}{x} = 4cos theta
$$
so we get
begin{align}
intfrac{sqrt{16x^2 – 9}}{x};dx &= int 4cos theta
frac{3}{4}cscthetacottheta;dtheta
\ &= 3intfrac{cos^2theta}{sin^2theta};dtheta =
3big(cot theta + thetabig)+C
end{align}
and then look at the picture to get
$$
3big(cot theta + thetabig)+C=
3 left[frac{sqrt{16x^23^2}}{3} + arcsinfrac{3}{4x}right]+C
$$
this method also works for “sum of squares”. Draw the right triangle
suggested by that particular sum of squares.
$$
16x^2 – 9 = 9left( left( tfrac{4x}{3} right)^2 – 1 right) = 9(sec^2theta – 1) = 9tan^2theta.
$$
I am confused by the suggestion to use trigonometric substitution, since $$frac{sqrt{16x^2 – 9}}{x} = 16 x frac{sqrt{16x^2 – 9}}{16x^2},$$ and the substitution $$u^2 = 16x^2 – 9, quad 2u , du = 32 x , dx$$ yields $$begin{align*}
int frac{sqrt{16x^2 – 9}}{x} , dx
&= int frac{u}{u^2 + 9} u , du \
&= int 1 – frac{9}{u^2 + 9} , du \
&= u – 3 tan^{1} frac{u}{3} + C \
&= sqrt{16x^2 – 9} – 3 tan^{1} frac{sqrt{16x^2 – 9}}{3} + C.
end{align*}$$
Trigonometric substitution certainly works, but in such cases, we can certainly avoid it.
Substitute $sec t= frac43 x$ to integrate
$$int frac{sqrt{16x^2 – 9}}{x} dx= 3int tan^2tdt= 3int( sec^2t 1)dt = 3tan t 3t+C
$$
For such radicals, I find the hyperbolic substitution easier.
With $dfrac43x=cosh t$,
$$intfrac{sqrt{16x^29}}xdx=3intfrac{sqrt{dfrac{16}9x^21}}xdx=3intfrac{sinh^2t}{cosh t}dt=3intfrac{dt}{cosh t}+3intcosh t,dt.$$
Given $int frac{sqrt{16x^2}9}{x}dx$ and that $sqrt{x^2a^2} Rightarrow x=a sec theta wedge a sec theta tan theta dtheta =dx$
Then,
$$int frac{sqrt{16x^2}9}{x}dx Rightarrow int frac{sqrt{16(3 sectheta)^2}3^2}{3 sec theta} 3 sec theta tan theta dtheta $$
$$ = 12int tan theta sqrt{sec^2 theta}9 $$
$$ = 12sqrt {sec^2theta} 81theta +C$$
by factoring out constants and integrating the sum term by term.
First get rid of the annoying factors,
$$intfrac{sqrt{16x^29}}xdx=intfrac{sqrt{16left(dfrac{3y}4right)^29}}{dfrac{3y}4}ddfrac{3y}4=3intfrac{sqrt{y^21}}{y}dy.$$
Then observe the identity
$$left(frac 12left(t+dfrac1tright)right)^21=left(frac 12left(tdfrac1tright)right)^2.$$
Then with $y=dfrac 12left(t+dfrac1tright)$ and $dy=dfrac 12left(1dfrac1{t^2}right)$,
$$intfrac{sqrt{y^21}}{y}dy=intfrac{dfrac 12left(tdfrac1tright)}{dfrac 12left(t+dfrac1tright)}dfrac 12left(1dfrac1{t^2}right)dt=frac12intleft(1+frac1{t^2}frac{4}{t^2+1}right) dt$$
which is easy.
For this problem, you goal with trig substitution to convert the radical into $sqrt{a(sec^2 x1)}$ which equals $tan x cdot sqrt{a}$.
Therefore, let $x=dfrac{3sec u}{4}$ so that when $x$ is squared in the radical, you’re left with $sqrt{9left(sec^2 u1right)}$.