MHB -7.3.89 Integral with trig subst

[karush]

$\begin{array}{lll}
I&=\displaystyle\int{\frac{dx}{x^2\sqrt{x^2-16}}}
\quad x=4\sec\theta
\quad dx=4\tan \theta\sec \theta
\end{array}$

just seeing if I started with the right x and dx or is there better

Mahalo

 

[topsquark]

Details! [math]dx = 4 ~ tan( \theta ) ~ sec( \theta ) ~ d \theta [/math].

Otherwise, good.

-Dan

 

[Prove It]

That substitution would be fine, but I would probably lean more towards a hyperbolic substitution (just personal preference).

$\displaystyle \begin{align*} x = 4\cosh{ \left( t \right) } \implies \mathrm{d}x = 4\sinh{\left( t \right) } \,\mathrm{d}t \end{align*}$

 

[karush]

so finally maybe...

Evaluate the Integral $I=\int{\dfrac{dx}{x^2\sqrt{x^2-16}}}
\quad x=4\sec\theta
\quad dx=4\tan \theta\sec \theta$
thus $\sec^{-1} \dfrac{x}{4} \sin\theta =\dfrac{\sqrt{x^2-16}}{x}$
Substitute
$\begin{array}{lll}
&=\int{\dfrac{4{{tan} \theta {\sec \theta }}}
{16{{\sec}^2 \theta \cdot 4{{tan} \theta }}}}\\
&=\dfrac{1}{16}\int\cos\theta \, d\theta
=\dfrac{\sin \theta}{16}\\
&=\dfrac{\dfrac{\sqrt{x^2-16}}{x}}{16}\\
&=\dfrac{\sqrt{x^2-16}}{16x}\\
&=\int\sin^6 x(1-\sin^2 x) \, dx\\
&=\cos(x) dx \\
&u=\sin{x}\ du=\cos x \, dx\\
&=\int{(u^6-u^8)\ du}\implies \dfrac{u^7}{7}-\dfrac{u^9}{9}+C\\
&=\dfrac{{{{sin}}^7 x\ }}{7}-\dfrac{{{{sin}}^9 x\ }}{9}+C
\end{array}$

 

[karush]

so finally maybe...

Evaluate the Integral $I=\int{\dfrac{dx}{x^2\sqrt{x^2-16}}}
\quad x=4\sec\theta
\quad dx=4\tan \theta\sec \theta$
thus $\sec^{-1} \dfrac{x}{4} \sin\theta =\dfrac{\sqrt{x^2-16}}{x}$
Substitute
$\displaystyle\int{\dfrac{4{{tan} \theta {\sec \theta }}}
{16{{\sec}^2 \theta \cdot 4{{tan} \theta }}}}d\theta$