`int x / sqrt(9 + 8x^2 - x^4) dx` Find or evaluate the integral by completing the square

Sunday, October 14, 2012

$\displaystyle\int\frac{{x}}{\sqrt{{{9}+{8}{{x}}^{{2}}-{{x}}^{{4}}}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

Recall that $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$ where:

f(x) as the integrand function

F(x) as the antiderivative of f(x)

C as the constant of integration..

For the given problem, the integral: $\displaystyle\int\frac{{x}}{\sqrt{{{9}+{8}{{x}}^{{2}}-{{x}}^{{4}}}}}{\left.{d}{x}\right.}$

does not yet resemble any formula from table of integrals.

To evaluate this, we are to apply u-substitution by letting:

$\displaystyle{u}={{x}}^{{2}}$ then $\displaystyle{{u}}^{{2}}={{x}}^{{4}}$ and $\displaystyle{d}{u}={2}{x}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{2}}={x}{\left.{d}{x}\right.}$ .

Then the integral becomes:

$\displaystyle\int\frac{{x}}{\sqrt{{{9}+{8}{{x}}^{{2}}-{{x}}^{{4}}}}}{\left.{d}{x}\right.}=\int{x}\frac{{\left.{d}{x}}}{\sqrt{{{9}+{8}{{x}}^{{2}}-{{x}}^{{4}}}}}$

$\displaystyle=\int\frac{{\frac{{{d}{u}}}{{2}}}}{\sqrt{{{9}+{8}{u}-{{u}}^{{4}}}}}$

Apply the basic property of integration: $\displaystyle\int{c}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ to factor out $\displaystyle\frac{{1}}{{2}}$ .

$\displaystyle\int\frac{{\frac{{{d}{u}}}{{2}}}}{\sqrt{{{9}+{8}{u}-{{u}}^{{4}}}}}=\frac{{1}}{{2}}\int\frac{{{d}{u}}}{\sqrt{{{9}+{8}{u}-{{u}}^{{4}}}}}$

The integral does not yet resembles any integration formula.

For further step, we apply completing the square on the part: $\displaystyle{9}+{8}{u}-{{u}}^{{2}}$ .

Completing the square:

Factoring out -1 from $\displaystyle{9}+{8}{u}-{{u}}^{{2}}$ becomes: $\displaystyle{\left(-{1}\right)}{\left(-{9}-{8}{{u}}^{{2}}+{{u}}^{{2}}\right)}$ or $\displaystyle-{\left({{u}}^{{2}}-{8}{u}-{9}\right)}$ .

$\displaystyle{{u}}^{{2}}-{8}{u}-{9}$ resembles $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}$ where:

$\displaystyle{a}={1}$ , $\displaystyle{b}=-{8}$ and $\displaystyle{c}={9}$ .

To complete the square we add and subtract $\displaystyle{{\left(-\frac{{b}}{{{2}{a}}}\right)}}^{{2}}$ .

Plug-in the value of $\displaystyle{a}={1}$ and $\displaystyle{b}=-{8}$ in $\displaystyle{{\left(-\frac{{b}}{{{2}{a}}}\right)}}^{{2}}$ :

$\displaystyle{{\left(-\frac{{b}}{{{2}{a}}}\right)}}^{{2}}={{\left(-\frac{{-{8}}}{{{2}\cdot{1}}}\right)}}^{{2}}$

$\displaystyle={{\left(\frac{{8}}{{2}}\right)}}^{{2}}$

$\displaystyle={{4}}^{{2}}$

$\displaystyle={16}.$

Adding and subtracting -16 inside the ():

$\displaystyle-{\left({{u}}^{{2}}-{8}{u}-{9}\right)}=-{\left({{u}}^{{2}}-{8}{u}-{9}+{16}-{16}\right)}$

To move out "-9" and "-16" outside the (), we distribute the negative sign or (-1).

$\displaystyle-{\left({{u}}^{{2}}-{8}{u}-{9}+{16}-{16}\right)}=-{\left({{u}}^{{2}}-{8}{u}-{9}+{16}\right)}+{\left(-{1}\right)}{\left(-{9}\right)}+{\left(-{1}\right)}{\left(-{16}\right)}$

$\displaystyle=-{\left({{u}}^{{2}}-{8}{u}-{9}+{16}\right)}+{9}+{16}$

$\displaystyle=-{\left({{u}}^{{2}}-{8}{u}-{9}+{16}\right)}+{25}$

Factor out the perfect square trinomial: $\displaystyle{{u}}^{{2}}-{8}{u}+{16}={{\left({u}-{4}\right)}}^{{2}}$

$\displaystyle-{\left({{u}}^{{2}}-{8}{u}+{16}\right)}+{16}=-{{\left({u}-{4}\right)}}^{{2}}+{25}$

Then it shows that $\displaystyle{9}+{8}{u}-{{u}}^{{4}}=-{{\left({u}-{4}\right)}}^{{2}}+{25}$

$\displaystyle={25}-{{\left({u}-{4}\right)}}^{{2}}$

$\displaystyle={{5}}^{{2}}-{{\left({u}-{4}\right)}}^{{2}}$

Then,

$\displaystyle\frac{{1}}{{2}}\int\frac{{{d}{u}}}{\sqrt{{{9}+{8}{u}-{{u}}^{{4}}}}}=\frac{{1}}{{2}}\int\frac{{{d}{u}}}{\sqrt{{{{5}}^{{2}}-{{\left({u}-{4}\right)}}^{{2}}}}}$

The integral part resembles the basic integration formula for inverse sine function:

$\displaystyle\int\frac{{{d}{u}}}{\sqrt{{{{a}}^{{2}}-{{u}}^{{2}}}}}={\arcsin{{\left(\frac{{u}}{{a}}\right)}}}+{C}$

Applying the formula, we get:

$\displaystyle\frac{{1}}{{2}}\int\frac{{{d}{u}}}{\sqrt{{{{5}}^{{2}}-{{\left({u}-{4}\right)}}^{{2}}}}}=\frac{{1}}{{2}}{\arcsin{{\left(\frac{{{u}-{4}}}{{5}}\right)}}}+{C}$

Plug-in $\displaystyle{u}={{x}}^{{2}}$ for the final answer:

$\displaystyle\int\frac{{x}}{\sqrt{{{9}+{8}{{x}}^{{2}}-{{x}}^{{4}}}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\arcsin{{\left(\frac{{{{x}}^{{2}}-{4}}}{{5}}\right)}}}+{C}$

Blog

Sunday, October 14, 2012

$\displaystyle\int\frac{{x}}{\sqrt{{{9}+{8}{{x}}^{{2}}-{{x}}^{{4}}}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

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?

Sunday, October 14, 2012

Find or evaluate the integral by completing the square

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?

$\displaystyle\int\frac{{x}}{\sqrt{{{9}+{8}{{x}}^{{2}}-{{x}}^{{4}}}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

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