The Notes: `int (dx)/[(ax)^2 - b^2]^(3/2)` Evaluate the integral

Tuesday, January 28, 2014

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{\left[{{\left({a}{x}\right)}}^{{2}}-{{b}}^{{2}}\right]}}^{{\frac{{3}}{{2}}}}}$ Evaluate the integral

$\displaystyle\int\frac{{\left.{d}{x}}}{{{\left[{{\left({a}{x}\right)}}^{{2}}-{{b}}^{{2}}\right]}}^{{\frac{{3}}{{2}}}}}$

Let's use the integral substitution,

Let u=ax

$\displaystyle{d}{u}={a}{\left.{d}{x}\right.}$

$\displaystyle\Rightarrow{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{a}}$

$\displaystyle=\int\frac{{{d}{u}}}{{{a}{{\left({{u}}^{{2}}-{{b}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}}}$

$\displaystyle=\frac{{1}}{{a}}\int\frac{{{d}{u}}}{{{\left({{u}}^{{2}}-{{b}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}}$

Now let's use the trigonometric substitution,

Let $\displaystyle{u}={b}{\sec{{\left(\theta\right)}}}$

so $\displaystyle{d}{u}={b}{\sec{{\left(\theta\right)}}}{\tan{{\left(\theta\right)}}}{d}\theta$

Plug these in the integrand,

$\displaystyle=\frac{{1}}{{a}}\int\frac{{{b}{\sec{{\left(\theta\right)}}}{\tan{{\left(\theta\right)}}}}}{{{\left({{b}}^{{2}}{{\sec}}^{{2}}{\left(\theta\right)}-{{b}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}}{d}\theta$

$\displaystyle=\frac{{1}}{{a}}\int\frac{{{b}{\sec{{\left(\theta\right)}}}{\tan{{\left(\theta\right)}}}}}{{{\left({{b}}^{{2}}{\left({{\sec}}^{{2}}{\left(\theta\right)}-{1}\right)}\right)}}^{{\frac{{3}}{{2}}}}}{d}\theta$

$\displaystyle=\frac{{1}}{{a}}\int\frac{{{b}{\sec{{\left(\theta\right)}}}{\tan{{\left(\theta\right)}}}}}{{{{\left({{b}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}{{\left({{\sec}}^{{2}}{\left(\theta\right)}-{1}\right)}}^{{\frac{{3}}{{2}}}}}}{d}\theta$

Now use the identity: $\displaystyle{{\tan}}^{{2}}{\left(\theta\right)}={{\sec}}^{{2}}{\left(\theta\right)}-{1}$

$\displaystyle=\frac{{1}}{{a}}\int\frac{{{b}{\sec{{\left(\theta\right)}}}{\tan{{\left(\theta\right)}}}}}{{{{b}}^{{3}}{{\left({{\tan}}^{{2}}{\left(\theta\right)}\right)}}^{{\frac{{3}}{{2}}}}}}{d}\theta$

$\displaystyle=\frac{{1}}{{a}}\int\frac{{{\sec{{\left(\theta\right)}}}{\tan{{\left(\theta\right)}}}}}{{{{b}}^{{2}}{{\tan}}^{{3}}{\left(\theta\right)}}}{d}\theta$

$\displaystyle=\frac{{1}}{{{a}{{b}}^{{2}}}}\int\frac{{\sec{{\left(\theta\right)}}}}{{{{\tan}}^{{2}}{\left(\theta\right)}}}{d}\theta$

$\displaystyle=\frac{{1}}{{{a}{{b}}^{{2}}}}\int\frac{{\frac{{1}}{{\cos{{\left(\theta\right)}}}}}}{{\frac{{{{\sin}}^{{2}}{\left(\theta\right)}}}{{{{\cos}}^{{2}}{\left(\theta\right)}}}}}{d}\theta$

$\displaystyle=\frac{{1}}{{{a}{{b}}^{{2}}}}\int{\left(\frac{{1}}{{\cos{{\left(\theta\right)}}}}\right)}\cdot\frac{{{{\cos}}^{{2}}{\left(\theta\right)}}}{{{{\sin}}^{{2}}{\left(\theta\right)}}}{d}\theta$

$\displaystyle=\frac{{1}}{{{a}{{b}}^{{2}}}}\int\frac{{\cos{{\left(\theta\right)}}}}{{{{\sin}}^{{2}}{\left(\theta\right)}}}{d}\theta$

Now let $\displaystyle{v}={\sin{{\left(\theta\right)}}}$

$\displaystyle\Rightarrow{d}{v}={\cos{{\left(\theta\right)}}}{d}\theta$

$\displaystyle=\frac{{1}}{{{a}{{b}}^{{2}}}}\int\frac{{1}}{{{v}}^{{2}}}{d}{v}$

$\displaystyle=\frac{{1}}{{{a}{{b}}^{{2}}}}{\left(\frac{{{v}}^{{-{2}+{1}}}}{{-{2}+{1}}}\right)}$

$\displaystyle=\frac{{1}}{{{a}{{b}}^{{2}}}}{\left(-\frac{{1}}{{v}}\right)}$

substitute back $\displaystyle{v}={\sin{{\left(\theta\right)}}}$

$\displaystyle=-\frac{{1}}{{{a}{{b}}^{{2}}{\sin{{\left(\theta\right)}}}}}$

We have used the substitution $\displaystyle{u}={b}{\sec{{\left(\theta\right)}}}$

So, $\displaystyle{\cos{{\left(\theta\right)}}}=\frac{{b}}{{u}}$

using pythagorean identity,

$\displaystyle{{\sin}}^{{2}}{\left(\theta\right)}+{{\cos}}^{{2}}{\left(\theta\right)}={1}$

$\displaystyle{{\sin}}^{{2}}{\left(\theta\right)}+{{\left(\frac{{b}}{{u}}\right)}}^{{2}}={1}$

$\displaystyle{{\sin}}^{{2}}{\left(\theta\right)}={1}-\frac{{{b}}^{{2}}}{{{u}}^{{2}}}$

$\displaystyle{{\sin}}^{{2}}{\left(\theta\right)}=\frac{{{{u}}^{{2}}-{{b}}^{{2}}}}{{{u}}^{{2}}}$

$\displaystyle{\sin{{\left(\theta\right)}}}=\frac{\sqrt{{{{u}}^{{2}}-{{b}}^{{2}}}}}{{u}}$

Also recall we have used u=ax,

$\displaystyle\therefore{\sin{{\left(\theta\right)}}}=\frac{\sqrt{{{{\left({a}{x}\right)}}^{{2}}-{{b}}^{{2}}}}}{{{a}{x}}}$

$\displaystyle=-\frac{{1}}{{{a}{{b}}^{{2}}\frac{\sqrt{{{{\left({a}{x}\right)}}^{{2}}-{{b}}^{{2}}}}}{{{a}{x}}}}}$

$\displaystyle={\left(-\frac{{1}}{{{{b}}^{{2}}}}\right)}{\left(\frac{{x}}{\sqrt{{{{\left({a}{x}\right)}}^{{2}}-{{b}}^{{2}}}}}\right)}$

Add a constant C to the solution,

$\displaystyle={\left(-\frac{{1}}{{{b}}^{{2}}}\right)}{\left(\frac{{x}}{\sqrt{{{{\left({a}{x}\right)}}^{{2}}-{{b}}^{{2}}}}}\right)}+{C}$

The Notes

Tuesday, January 28, 2014

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{{{\left[{{\left({a}{x}\right)}}^{{2}}-{{b}}^{{2}}\right]}}^{{\frac{{3}}{{2}}}}}$ Evaluate the integral

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