Trig Substitution for Whitman 8.4.8: Complete the Square

[karush]

Whitman 8.4.8 Trig substitution?

Whitman 8.4.8
Complete the square..
\begin{align*}
\int\sqrt{x^{2}-2x}dx &=\int\sqrt{x^{2}-2x+1-1}dx\\
&=\int\sqrt{(x-1)^{2}-1^{2}}dx\\
&=\int\sqrt{U^{2}-1^{2}}dx\\
\end{align*}
Was wondering what substation best to use

Can $\cosh\theta \text{ or } \sinh\theta \text{ be used? } $

The table will give this but don't know how it was derived?

$$\int\sqrt{x^{2}-a^{2}}dx
=\frac{x}{2}\sqrt{x^{2}-a^{2}}
-\frac{a^{2}}{2}\log |x
+\sqrt{x^{2}-a^{2}}|+c.$$
So..
$$=\frac{x}{2}\sqrt{(x-1)^{2}-1}
-\frac{1}{2}\log |x
+\sqrt{(x-1)^{2}-1}|
+c.$$

 

[MarkFL]

A hyperbolic trig. identity we can use is:

\(\displaystyle \cosh^2(z)-\sinh^2(z)=1\)

How would you proceed from here? :)

 

[karush]

Whitman 8.4.8
Complete the square..
\begin{align*}
\int\sqrt{x^{2}-2x}dx &=\int\sqrt{x^{2}-2x+1-1}dx\\
&=\int\sqrt{(x-1)^{2}-1^{2}}dx\\
&=\int\sqrt{u^{2}-1^{2}}du\\
\end{align*}
$$ \sinh^2{x}=\cosh^2 {x}-1$$
$u=\cosh{x} \ \ \ du=\sinh{x} dx$
Then
$\displaystyle
\int\sqrt{\cosh^2 {u} -1^{2}}du
=\int \sinh^2 {u} \ du$

So..
$$=\frac{x}{2}\sqrt{(x-1)^{2}-1}
-\frac{1}{2}\log |x
+\sqrt{(x-1)^{2}-1}|
+c.$$

 

[MarkFL]

Okay, we are given:

\(\displaystyle I=\int \sqrt{x^2-2x}\,dx\)

And completing the square:

\(\displaystyle I=\int \sqrt{(x-1)^2-1}\,dx\)

Now, at this point you want to make the substitution:

\(\displaystyle x-1=\cosh(u)\,\therefore\,dx=\sinh(u)\,du\)

And we have:

\(\displaystyle I=\int \sqrt{\cosh^2(u)-1}\,\sinh(u)\,du=\int \sinh^2(u)\,du\)

Now you want to use the identity:

\(\displaystyle \cosh(2z)=2\sinh^2(z)+1\)

 

[karush]

$$\displaystyle \frac{\cosh{2z}}{2}
-\frac{1}{2}
=\sinh^2{z}$$
So
$$\frac{1}{2}\int \cosh{2z}\
- \int \frac{1}{2} \ dx $$
Sorry lots of @#$ with WiFi here

 

[MarkFL]

$$\displaystyle \frac{\cosh{2z}}{2}
-\frac{1}{2}
=\sinh^2{z}$$
So
$$\frac{1}{2}\int \cosh{2z}\
- \int \frac{1}{2} \ dx $$
Sorry lots of @#$ with WiFi here

Well, I would just write the next step as:

\(\displaystyle I=\frac{1}{2}\int \cosh(2u)-1\,du\)

Now, it is fairly simple to obtain the anti-derivative...then you go through the process of back-substituting for $u$.

 

[karush]

$\displaystyle I=
\frac{1}{2}\int \cosh(2u)-1\,du=
\frac{e^{2u}\left(e^{4u}-4ue^{2u}-1\right)}{8}+C$

 

[MarkFL]

I was thinking more along the lines of:

\(\displaystyle I=\frac{1}{4}\sinh(2u)-\frac{1}{2}u+C\)

Next, use the identity:

\(\displaystyle \sinh(2z)=2\sinh(z)\cosh(z)\):

\(\displaystyle I=\frac{1}{2}\sinh(u)\cosh(u)-\frac{1}{2}u+C\)

Next, recall we used:

\(\displaystyle \cosh(u)=x-1\)

\(\displaystyle \sinh(u)=\sqrt{\cosh^2(u)-1}=\sqrt{(x-1)^2-1}\)

And let's also apply the inverse function as a log formula:

\(\displaystyle \arcosh(z)=\ln\left(z+\sqrt{z^2-1}\right)\)

To state:

\(\displaystyle I=\frac{1}{2}\left(\sqrt{(x-1)^2-1}\right)(x-1)-\frac{1}{2}\ln\left|x-1+\sqrt{(x-1)^2-1}\right|+C\)

Now, observing the first term, when expanded, contains a constant, we may write:

\(\displaystyle I=\frac{x}{2}\sqrt{x^2-2x}-\frac{1}{2}\ln\left|x-1+\sqrt{x(x-2)}\right|+C\)

Next, applying some log properties:

\(\displaystyle I=\frac{x}{2}\sqrt{x^2-2x}-\frac{1}{2}\ln\left|2(x-1)+2\sqrt{x(x-2)}\right|+C+\frac{1}{2}\ln(2)\)

\(\displaystyle I=\frac{x}{2}\sqrt{x^2-2x}-\frac{1}{2}\ln\left|x+2\sqrt{x(x-2)}+x-2\right|+C\)

\(\displaystyle I=\frac{x}{2}\sqrt{x^2-2x}-\frac{1}{2}\ln\left|\left(\sqrt{x}+\sqrt{x-2}\right)^2\right|+C\)

\(\displaystyle I=\frac{x}{2}\sqrt{x^2-2x}-\ln\left|\sqrt{x}+\sqrt{x-2}\right|+C\)

 

[karush]

Wow. That was great help.
Frankly I would of derailed inspite of all the other examples I looked at

Just glad I'm getting all the great MHB help now
before hitting it all cold when
the Calc II class starts up in August

I can see now how these substitutions work.

I'll keep trying more of these