Help Notes: `int sqrt(1-x^2)/x^4 dx` Find the indefinite integral

Monday, 15 April 2013

$\displaystyle\int\frac{\sqrt{{{1}-{{x}}^{{2}}}}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$ Find the indefinite integral

Given
,

$\displaystyle\int\frac{\sqrt{{{1}-{{x}}^{{2}}}}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$

By applying Integration by
parts we can solve the given integral

so,

let $\displaystyle{u}=$
sqrt(1-x^2) ,v' = (1/x^4)

=> $\displaystyle{u}'={\left(\sqrt{{{1}-{{x}}^{{2}}}}\right)}'$

=> = $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left(\sqrt{{{1}-{{x}}^{{2}}}}\right)}$

let $\displaystyle{t}={1}-{{x}}^{{2}}$

so,

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left(\sqrt{{{1}-{{x}}^{{2}}}}\right)}$

= $\displaystyle\frac{{d}}{{\left.{d}{x}}}$
(sqrt(t))

= $\displaystyle\frac{{d}}{{{\left.{d}{t}\right.}}}\sqrt{{{t}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({t}\right)}$ [as $\displaystyle\frac{{d}}{{\left.{d}{x}}}{f{{\left({t}\right)}}}=\frac{{d}}{{{\left.{d}{t}\right.}}}{f{{\left({t}\right)}}}$
(dt)/dx $\displaystyle{\]}$

= $\displaystyle{\left[{\left(\frac{{1}}{{2}}\right)}{{t}}^{{{\left(\frac{{1}}{{2}}\right)}-{1}}}\right]}\cdot{\left(\frac{{d}}{{\left.{d}{x}}}{\left({1}-{{x}}^{{2}}\right)}\right)}$

=
$\displaystyle{\left[{\left(\frac{{1}}{{2}}\right)}{{t}}^{{-\frac{{1}}{{2}}}}\right]}\cdot{\left(-{2}{x}\right)}$

= $\displaystyle{\left[\frac{{1}}{{{2}\sqrt{{{1}-{{x}}^{{2}}}}}}\right]}\cdot{\left(-{2}{x}\right)}$

= $\displaystyle-\frac{{x}}{\sqrt{{{1}-{{x}}^{{2}}}}}$

so, $\displaystyle{u}'=-\frac{{x}}{\sqrt{{{1}-{{x}}^{{2}}}}}$ and as $\displaystyle{v}'={\left(\frac{{1}}{{{x}}^{{4}}}\right)}$
so

$\displaystyle{v}=\int\frac{{1}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$

= $\displaystyle\int{{x}}^{{-{4}}}{\left.{d}{x}\right.}$

= $\displaystyle\frac{{{{x}}^{{-{4}+{1}}}}}{{-{4}+{1}}}$

= $\displaystyle\frac{{{{x}}^{{-{3}}}}}{{-{3}}}$

=
$\displaystyle-{\left(\frac{{1}}{{{3}{{x}}^{{3}}}}\right)}$

so , let us see the values
altogether.

$\displaystyle{u}=\sqrt{{{1}-{{x}}^{{2}}}},{u}'=-\frac{{x}}{\sqrt{{{1}-{{x}}^{{2}}}}}$ and $\displaystyle{v}'={\left(\frac{{1}}{{{x}}^{{4}}}\right)}$
,v=-(1/(3x^3))

Now ,by applying the integration by
parts $\displaystyle\int{u}{v}'$ is given as

$\displaystyle\int{u}{v}'={u}{v}-\int{u}'{v}$

then,

$\displaystyle\int\frac{\sqrt{{{1}-{{x}}^{{2}}}}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\sqrt{{{1}-{{x}}^{{2}}}}\right)}$
(-(1/(3x^3))) - int (-x/sqrt(1-x^2))(-(1/(3x^3))) dx

= $\displaystyle{\left(\sqrt{{{1}-{{x}}^{{2}}}}\right)}$
(-(1/(3x^3))) -(-...

Help Notes

Monday, 15 April 2013

$\displaystyle\int\frac{\sqrt{{{1}-{{x}}^{{2}}}}}{{{x}}^{{4}}}{\left.{d}{x}\right.}$ Find the indefinite integral

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