Steward e6 {7r11} Integral substitution

[karush]

$\tiny\text{Steward e6 {7r11} } $
$$\displaystyle
I=\int_1^2 {\frac{\sqrt{{x}^{2}-1}}{x} } \ d{x} =
{\sqrt{3}+\frac{1}{3}\pi} $$
Again what substitution was the question but tried..
$$\begin{align}
u& = {x}^{2}-1 &
du&= 2x \ d{t} &
x&=\sqrt{u+1}
\end{align}$$
$$\displaystyle
I=\int_1^2 \frac{\sqrt{u}}{\sqrt{u+1}} \ d{u}$$
$$\begin{align}
u& = \tan^2\left({w}\right) &
du&=\frac{2\sin\left({w}\right)}{\cos^3\left({w}\right)} \ d{w} &
w&=\arctan{\sqrt{u}+\pi}
\end{align}$$
Proceed ?$\tiny\text{from Surf the Nations math study group}$

 

[MarkFL]

This is how I would work the problem:

\(\displaystyle I=\int_1^2 \frac{\sqrt{x^2-1}}{x}\,dx=\int_1^2 \sqrt{1-x^{-2}}\,dx\)

Let:

\(\displaystyle x^{-1}=\sin(u)\implies \,dx=-\frac{\cos(u)}{\sin^2(u)}\,du\)

And we now have:

\(\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cot^2(u)\,du=-\left[u+\cot(u)\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}=\frac{\pi}{6}+\sqrt{3}-\frac{\pi}{2}-0=\sqrt{3}-\frac{\pi}{3}\)

 

[karush]

I would have never thought of that..
Cool trick indeed!

 

[Prove It]

$\tiny\text{Steward e6 {7r11} } $
$$\displaystyle
I=\int_1^2 {\frac{\sqrt{{x}^{2}-1}}{x} } \ d{x} =
{\sqrt{3}+\frac{1}{3}\pi} $$
Again what substitution was the question but tried..
$$\begin{align}
u& = {x}^{2}-1 &
du&= 2x \ d{t} &
x&=\sqrt{u+1}
\end{align}$$
$$\displaystyle
I=\int_1^2 \frac{\sqrt{u}}{\sqrt{u+1}} \ d{u}$$
$$\begin{align}
u& = \tan^2\left({w}\right) &
du&=\frac{2\sin\left({w}\right)}{\cos^3\left({w}\right)} \ d{w} &
w&=\arctan{\sqrt{u}+\pi}
\end{align}$$
Proceed ?$\tiny\text{from Surf the Nations math study group}$

Your approach is fine, but you need to fiddle with the integrand first.

$\displaystyle \begin{align*} \int_1^2{\frac{\sqrt{x^2 - 1}}{x}\,\mathrm{d}x} &= \int_1^2{ \frac{2\,x\,\sqrt{x^2 - 1}}{2\,x^2}\,\mathrm{d}x } \end{align*}$

so let $\displaystyle \begin{align*} u = x^2 - 1 \implies \mathrm{d}u = 2\,x\,\mathrm{d}x \end{align*}$ with $\displaystyle \begin{align*} u(1) = 0 \end{align*}$ and $\displaystyle \begin{align*} u(2) = 3 \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int_1^2{ \frac{2\,x\,\sqrt{x^2 - 1}}{2\,x^2}\,\mathrm{d}x } &= \int_0^3{ \frac{\sqrt{u}}{2\,\left( u + 1 \right) }\,\mathrm{d}u } \end{align*}$

Now how might you continue?

 

[karush]

$\displaystyle \begin{align*} \int_1^2{ \frac{2\,x\,\sqrt{x^2 - 1}}{2\,x^2}\,\mathrm{d}x } &= \int_0^3{ \frac{\sqrt{u}}{2\,\left( u + 1 \right) }\,\mathrm{d}u } \end{align*} $

$\begin{align}\displaystyle
u& = \tan^2 \left({w}\right)&
du&=\frac{2\sin\left({w}\right)}{\cos^3\left({w}\right)} \ d{t} \\
\end{align}$
Then
$\displaystyle
I= \int_0^3{ \frac{\sqrt{u}}{2\,\left( u + 1 \right) }\,\mathrm{d}u } \\
\implies \frac{1 }{2}\int_0^3
{ \frac{\sqrt{\tan^2 \left({w}\right)}}
{\left( \tan^2 \left({w}\right) + 1 \right) }\,\mathrm{d}w } \\
\implies \frac{1 }{2} \int_0^3 \frac{\tan\left({w}\right)}
{\sec^2 \left({w}\right)}
\frac{2\sin\left({w}\right)}{\cos^3\left({w}\right)} \ dw $

Assume ok

 

[Greg]

\(\displaystyle u=\tan^2(w),\quad du=2\tan(w)\sec^2(w)\,dw\) (chain rule)

\(\displaystyle \int_0^3\dfrac{\sqrt u}{2(u+1)}\,du=\int_0^{\pi/3}\tan^2(w)\,dw=\int_0^{\pi/3}\sec^2(w)-1\,dw=\left[\tan(w)-w\right]_0^{\pi/3}=\sqrt{3}-\dfrac{\pi}{3}\)

 

[Prove It]

My approach:

$\displaystyle \begin{align*} \int_0^3{ \frac{\sqrt{u}}{2\,\left( u + 1 \right) } \,\mathrm{d}u } &= \int_0^3{ \frac{\left( \sqrt{u} \right) ^2 }{ \left[ \left( \sqrt{u} \right) ^2 + 1 \right] }\,\frac{1}{2\,\sqrt{u}}\,\mathrm{d}u} \end{align*}$

Now let $\displaystyle \begin{align*} v = \sqrt{u} \implies \mathrm{d}v = \frac{1}{2\,\sqrt{u}}\,\mathrm{d}u \end{align*}$ noting that $\displaystyle \begin{align*} v(0) = 0 \end{align*}$ and $\displaystyle \begin{align*} v(3) = \sqrt{3} \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int_0^3{ \frac{\left( \sqrt{u} \right) ^2}{\left[ \left( \sqrt{u} \right) ^2 + 1 \right] } \,\frac{1}{2\,\sqrt{u}}\,\mathrm{d}u } &= \int_0^{\sqrt{3}}{ \frac{v^2}{v^2 + 1}\,\mathrm{d}v } \\ &= \int_0^{\sqrt{3}}{ \left( 1 - \frac{1}{v^2 + 1} \right) \,\mathrm{d}v } \\ &= \left[ v - \arctan{\left( v \right) } \right] _0^{\sqrt{3}} \\ &= \left[ 0 - \arctan{(0)} \right] - \left[ \sqrt{3} - \arctan{ \left( \sqrt{3} \right) } \right] \\ &= \frac{\pi}{3} - \sqrt{3} \end{align*}$