`int_(sqrt(2)/3)^(2/3) (dx)/(x^5 sqrt(9x^2 - 1))` Evaluate the integral

Wednesday, April 29, 2015

$\displaystyle{\int_{{\frac{\sqrt{{{2}}}}{{3}}}}^{{\frac{{2}}{{3}}}}}\frac{{{\left.{d}{x}\right.}}}{{{{x}}^{{5}}\sqrt{{{9}{{x}}^{{2}}-{1}}}}}$ Evaluate the integral

$\displaystyle{\int_{{\frac{\sqrt{{{2}}}}{{3}}}}^{{\frac{{2}}{{3}}}}}\frac{{1}}{{{{x}}^{{5}}\sqrt{{{9}{{x}}^{{2}}-{1}}}}}{\left.{d}{x}\right.}$

Let's first evaluate the indefinite integral by integral substitution,

Let $\displaystyle{x}=\frac{{1}}{{3}}{\sec{{\left({u}\right)}}}$

$\displaystyle\Rightarrow{\left.{d}{x}\right.}=\frac{{1}}{{3}}{\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}{d}{u}$

$\displaystyle\int\frac{{1}}{{{{x}}^{{5}}\sqrt{{{9}{{x}}^{{2}}-{1}}}}}{\left.{d}{x}\right.}=\int{\left(\frac{{1}}{{{{\left(\frac{{1}}{{3}}{\sec{{\left({u}\right)}}}\right)}}^{{5}}\sqrt{{{9}{{\left(\frac{{1}}{{3}}{\sec{{\left({u}\right)}}}\right)}}^{{2}}-{1}}}}}\right)}\frac{{1}}{{3}}{\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}{d}{u}$

$\displaystyle=\int{\left(\frac{{1}}{{\frac{{1}}{{243}}{{\sec}}^{{5}}{\left({u}\right)}\sqrt{{{{\sec}}^{{2}}{\left({u}\right)}-{1}}}}}\right)}\frac{{1}}{{3}}{\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}{d}{u}$

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

$\displaystyle=\int{\left(\frac{{243}}{{{3}{{\sec}}^{{5}}{\left({u}\right)}\sqrt{{{1}+{{\tan}}^{{2}}{\left({u}\right)}-{1}}}}}\right)}{\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}{d}{u}$

$\displaystyle=\int\frac{{{81}{\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}}}{{{{\sec}}^{{5}}{\left({u}\right)}\sqrt{{{{\tan}}^{{2}}{\left({u}\right)}}}}}{d}{u}$

$\displaystyle={81}\int\frac{{1}}{{{{\sec}}^{{4}}{\left({u}\right)}}}{d}{u}$

$\displaystyle={81}\int{{\cos}}^{{4}}{\left({u}\right)}{d}{u}$

Now let's use the identity: $\displaystyle{{\cos}}^{{2}}{\left(\theta\right)}=\frac{{{1}+{\cos{{\left({2}\theta\right)}}}}}{{2}}$

$\displaystyle={81}\int{{\left(\frac{{{1}+{\cos{{\left({2}{u}\right)}}}}}{{2}}\right)}}^{{2}}{d}{u}$

$\displaystyle={81}\int\frac{{{1}+{{\cos}}^{{2}}{\left({2}{u}\right)}+{2}{\cos{{\left({2}{u}\right)}}}}}{{4}}{d}{u}$

$\displaystyle={81}\int{\left(\frac{{1}}{{4}}+\frac{{{{\cos}}^{{2}}{\left({2}{u}\right)}}}{{4}}+\frac{{1}}{{2}}{\cos{{\left({2}{u}\right)}}}\right)}{d}{u}$

$\displaystyle={81}{\left(\int\frac{{1}}{{4}}{d}{u}+\int\frac{{{{\cos}}^{{2}}{\left({2}{u}\right)}}}{{4}}{d}{u}+\int\frac{{{\cos{{\left({2}{u}\right)}}}}}{{2}}{d}{u}\right)}$

$\displaystyle={81}{\left(\frac{{u}}{{4}}+\frac{{1}}{{4}}\int\frac{{{1}+{\cos{{\left({4}{u}\right)}}}}}{{2}}{d}{u}+\frac{{1}}{{2}}\int{\cos{{\left({2}{u}\right)}}}{d}{u}\right)}$

$\displaystyle={81}{\left(\frac{{u}}{{4}}+\frac{{1}}{{4}}\int{\left(\frac{{1}}{{2}}+\frac{{\cos{{\left({4}{u}\right)}}}}{{2}}\right)}{d}{u}+\frac{{1}}{{2}}\frac{{{\sin{{\left({2}{u}\right)}}}}}{{2}}\right)}$

$\displaystyle={81}{\left(\frac{{u}}{{4}}+\frac{{1}}{{4}}{\left(\int\frac{{1}}{{2}}{d}{u}+\int\frac{{\cos{{\left({4}{u}\right)}}}}{{2}}{d}{u}\right)}+\frac{{1}}{{4}}{\sin{{\left({2}{u}\right)}}}\right)}$

$\displaystyle={81}{\left(\frac{{u}}{{4}}+\frac{{1}}{{4}}{\left(\frac{{u}}{{2}}+\frac{{1}}{{2}}\frac{{\sin{{\left({4}{u}\right)}}}}{{4}}\right)}+\frac{{1}}{{4}}{\sin{{\left({2}{u}\right)}}}\right)}$ $\displaystyle={81}{\left(\frac{{u}}{{4}}+\frac{{u}}{{8}}+\frac{{\sin{{\left({4}{u}\right)}}}}{{32}}+\frac{{\sin{{\left({2}{u}\right)}}}}{{4}}\right)}$

$\displaystyle={81}{\left(\frac{{{3}{u}}}{{8}}+\frac{{\sin{{\left({4}{u}\right)}}}}{{32}}+\frac{{\sin{{\left({2}{u}\right)}}}}{{4}}\right)}$

Now recall that we have used $\displaystyle{x}=\frac{{1}}{{3}}{\sec{{\left({u}\right)}}}$

$\displaystyle\Rightarrow{\sec{{\left({u}\right)}}}={3}{x}$

$\displaystyle\Rightarrow{\cos{{\left({u}\right)}}}=\frac{{1}}{{{3}{x}}}$

$\displaystyle\Rightarrow{u}={\arccos{{\left(\frac{{1}}{{{3}{x}}}\right)}}}$

Substitute back u and add a constant C to the solution,

$\displaystyle={81}{\left(\frac{{3}}{{8}}{\arccos{{\left(\frac{{1}}{{{3}{x}}}\right)}}}+\frac{{1}}{{32}}{\sin{{\left({4}{\arccos{{\left(\frac{{1}}{{{3}{x}}}\right)}}}\right)}}}+\frac{{1}}{{4}}{\sin{{\left({2}{\arccos{{\left(\frac{{1}}{{{3}{x}}}\right)}}}\right)}}}\right)}+{C}$

Now let's evaluate the definite integral,

$\displaystyle{\int_{{\frac{\sqrt{{{2}}}}{{3}}}}^{{\frac{{2}}{{3}}}}}\frac{{\left.{d}{x}}}{{{{x}}^{{5}}\sqrt{{{9}{{x}}^{{2}}-{1}}}}}={81}{{\left[\frac{{3}}{{8}}{\arccos{{\left(\frac{{1}}{{{3}{x}}}\right)}}}+\frac{{1}}{{32}}{\sin{{\left({4}{\arccos{{\left(\frac{{1}}{{{3}{x}}}\right)}}}\right)}}}+\frac{{1}}{{4}}{\sin{{\left({2}{\arccos{{\left(\frac{{1}}{{{3}{x}}}\right)}}}\right)}}}\right]}_{{\frac{\sqrt{{{2}}}}{{3}}}}^{{\frac{{2}}{{3}}}}}$

$\displaystyle={81}{\left[\frac{{3}}{{8}}{\arccos{{\left(\frac{{1}}{{2}}\right)}}}+\frac{{1}}{{32}}{\sin{{\left({4}{\arccos{{\left(\frac{{1}}{{2}}\right)}}}\right)}}}+\frac{{1}}{{4}}{\sin{{\left({2}{\arccos{{\left(\frac{{1}}{{2}}\right)}}}\right)}}}\right]}-{81}{\left[\frac{{3}}{{8}}{\arccos{{\left(\frac{{1}}{\sqrt{{{2}}}}\right)}}}+\frac{{1}}{{32}}{\sin{{\left({4}{\arccos{{\left(\frac{{1}}{\sqrt{{{2}}}}\right)}}}\right)}}}+\frac{{1}}{{4}}{\sin{{\left({2}{\arccos{{\left(\frac{{1}}{\sqrt{{{2}}}}\right)}}}\right)}}}\right]}$

$\displaystyle={81}{\left[\frac{{3}}{{8}}\cdot\frac{\pi}{{3}}+\frac{{1}}{{32}}{\sin{{\left({4}\cdot\frac{\pi}{{3}}\right)}}}+\frac{{1}}{{4}}{\sin{{\left({2}\cdot\frac{\pi}{{3}}\right)}}}\right]}-{81}{\left[\frac{{3}}{{8}}\cdot\frac{\pi}{{4}}+\frac{{1}}{{32}}{\sin{{\left({4}\cdot\frac{\pi}{{4}}\right)}}}+\frac{{1}}{{4}}{\sin{{\left({2}\cdot\frac{\pi}{{4}}\right)}}}\right]}$

$\displaystyle={81}{\left[\frac{\pi}{{8}}+\frac{{1}}{{32}}{\sin{{\left(\frac{{{4}\pi}}{{3}}\right)}}}+\frac{{1}}{{4}}{\sin{{\left(\frac{{{2}\pi}}{{3}}\right)}}}\right]}-{81}{\left[\frac{{{3}\pi}}{{32}}+\frac{{1}}{{32}}{\sin{{\left(\pi\right)}}}+\frac{{1}}{{4}}{\sin{{\left(\frac{\pi}{{2}}\right)}}}\right]}$

$\displaystyle={81}{\left[\frac{\pi}{{8}}+\frac{{1}}{{32}}{\left(-\frac{\sqrt{{{3}}}}{{2}}\right)}+\frac{{1}}{{4}}{\left(\frac{\sqrt{{{3}}}}{{2}}\right)}\right]}-{81}{\left[\frac{{{3}\pi}}{{32}}+\frac{{1}}{{32}}{\left({0}\right)}+\frac{{1}}{{4}}{\left({1}\right)}\right]}$

$\displaystyle={81}{\left[\frac{\pi}{{8}}+\frac{\sqrt{{{3}}}}{{2}}{\left(-\frac{{1}}{{32}}+\frac{{1}}{{4}}\right)}-\frac{{{3}\pi}}{{32}}-\frac{{1}}{{4}}\right]}$

$\displaystyle={81}{\left[\frac{\pi}{{8}}-\frac{{{3}\pi}}{{32}}+\frac{\sqrt{{{3}}}}{{2}}{\left(\frac{{7}}{{32}}\right)}-\frac{{1}}{{4}}\right]}$

$\displaystyle={81}{\left[\frac{\pi}{{32}}+\frac{{{7}\sqrt{{{3}}}}}{{64}}-\frac{{1}}{{4}}\right]}$

Blog

Wednesday, April 29, 2015

$\displaystyle{\int_{{\frac{\sqrt{{{2}}}}{{3}}}}^{{\frac{{2}}{{3}}}}}\frac{{{\left.{d}{x}\right.}}}{{{{x}}^{{5}}\sqrt{{{9}{{x}}^{{2}}-{1}}}}}$ Evaluate the integral

No comments:

Post a Comment

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Wednesday, April 29, 2015

Evaluate the integral

No comments:

Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{\int_{{\frac{\sqrt{{{2}}}}{{3}}}}^{{\frac{{2}}{{3}}}}}\frac{{{\left.{d}{x}\right.}}}{{{{x}}^{{5}}\sqrt{{{9}{{x}}^{{2}}-{1}}}}}$ Evaluate the integral

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?