New Notes Blog: `int sqrt(5x^2-1)dx` Find the indefinite integral

Tuesday, 8 October 2013

$\displaystyle\int\sqrt{{{5}{{x}}^{{2}}-{1}}}{\left.{d}{x}\right.}$ Find the indefinite integral

Recall that indefinite integral follows the formula: $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where: $\displaystyle{f{{\left({x}\right)}}}$ as the integrand

$\displaystyle{F}{\left({x}\right)}$ as the anti-derivative function

$\displaystyle{C}$ as the arbitrary constant known as constant of integration

For the given problem $\displaystyle\int\sqrt{{{5}{{x}}^{{2}}-{1}}}{\left.{d}{x}\right.}$ , it resembles one of the formula from integration table. We may apply the integral formula for function with roots as:

$\displaystyle\int\sqrt{{{{u}}^{{2}}\pm{{a}}^{{2}}}}{d}{u}=\frac{{1}}{{2}}{u}\sqrt{{{{u}}^{{2}}\pm{{a}}^{{2}}}}\pm\frac{{1}}{{2}}{{a}}^{{2}}{\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}\pm{{a}}^{{2}}}}\right|}+{C}$ .

Take...

Recall that indefinite integral follows the formula: $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where: $\displaystyle{f{{\left({x}\right)}}}$ as the integrand

$\displaystyle{F}{\left({x}\right)}$ as the anti-derivative function

$\displaystyle{C}$ as the arbitrary constant known as constant of integration

Take note the sign inside the root is " $\displaystyle{\left(-\right)}$ " then we follow the formula as:

$\displaystyle\int\sqrt{{{{u}}^{{2}}-{{a}}^{{2}}}}{d}{u}=\frac{{1}}{{2}}{u}\sqrt{{{{u}}^{{2}}-{{a}}^{{2}}}}-\frac{{1}}{{2}}{{a}}^{{2}}{\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}-{{a}}^{{2}}}}\right|}+{C}$

By comparing " $\displaystyle{{u}}^{{2}}-{{a}}^{{2}}$ " with " $\displaystyle{5}{{x}}^{{2}}-{1}$ " , we determine the corresponding values as:

$\displaystyle{{u}}^{{2}}={5}{{x}}^{{2}}$ or $\displaystyle{{\left(\sqrt{{{5}}}{x}\right)}}^{{2}}$ then $\displaystyle{u}=\sqrt{{{5}}}{x}$

$\displaystyle{{a}}^{{2}}={1}$ or $\displaystyle{{1}}^{{2}}$ then $\displaystyle{a}={1}$

For the derivative of $\displaystyle{u}$ , we get $\displaystyle{d}{u}=\sqrt{{{5}}}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{\sqrt{{{5}}}}={\left.{d}{x}\right.}$ .

Plug-in on the values $\displaystyle{{u}}^{{2}}={5}{{x}}^{{2}}$ and $\displaystyle\frac{{{d}{u}}}{\sqrt{{{5}}}}={\left.{d}{x}\right.}$ on the integral problem, we get:

$\displaystyle\int\sqrt{{{5}{{x}}^{{2}}-{1}}}{\left.{d}{x}\right.}=\int\sqrt{{{{u}}^{{2}}-{1}}}\cdot\frac{{{d}{u}}}{\sqrt{{{5}}}}$

Apply the basic properties of integration: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle\int\sqrt{{{{u}}^{{2}}-{1}}}\cdot\frac{{{d}{u}}}{\sqrt{{{5}}}}=\frac{{1}}{\sqrt{{{5}}}}\int\sqrt{{{{u}}^{{2}}-{1}}}{d}{u}$

Apply aforementioned integral formula for function with roots where $\displaystyle{{a}}^{{2}}={1}$ , we get:

$\displaystyle\frac{{1}}{\sqrt{{{5}}}}\int\sqrt{{{{u}}^{{2}}-{1}}}{d}{u}=\frac{{1}}{\sqrt{{{5}}}}\cdot{\left[\frac{{1}}{{2}}{u}\sqrt{{{{u}}^{{2}}-{1}}}-\frac{{1}}{{2}}\cdot{1}\cdot{\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}-{1}}}\right|}\right]}+{C}$

$\displaystyle=\frac{{1}}{\sqrt{{{5}}}}\cdot{\left[\frac{{1}}{{2}}{u}\sqrt{{{{u}}^{{2}}-{1}}}-\frac{{1}}{{2}}{\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}-{1}}}\right|}\right]}+{C}$

$\displaystyle=\frac{{1}}{{{2}\sqrt{{{5}}}}}{u}\sqrt{{{{u}}^{{2}}-{1}}}-\frac{{1}}{{{2}\sqrt{{{5}}}}}{\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}-{1}}}\right|}{\]}+{C}$

$\displaystyle=\frac{{{u}\sqrt{{{{u}}^{{2}}-{1}}}}}{{{2}\sqrt{{{5}}}}}-\frac{{{\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}-{1}}}\right|}}}{{{2}\sqrt{{{5}}}}}+{C}$

Plug-in $\displaystyle{{u}}^{{2}}={5}{{x}}^{{2}}$ and $\displaystyle{u}=\sqrt{{{5}}}{x}$ on $\displaystyle\frac{{{u}\sqrt{{{{u}}^{{2}}-{1}}}}}{{{2}\sqrt{{{5}}}}}-\frac{{{\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}-{1}}}\right|}}}{{{2}\sqrt{{{5}}}}}+{C}$ , we get the indefinite integral as:

$\displaystyle\int\sqrt{{{5}{{x}}^{{2}}-{1}}}{\left.{d}{x}\right.}=\frac{{\sqrt{{{5}}}{x}\sqrt{{{5}{{x}}^{{2}}-{1}}}}}{{{2}\sqrt{{{5}}}}}-\frac{{{\ln}{\left|\sqrt{{{5}}}{x}+\sqrt{{{5}{{x}}^{{2}}-{1}}}\right|}}}{{{2}\sqrt{{{5}}}}}+{C}$

$\displaystyle=\frac{{{x}\sqrt{{{5}{{x}}^{{2}}-{1}}}}}{{2}}-\frac{{{\ln}{\left|\sqrt{{{5}}}{x}+\sqrt{{{5}{{x}}^{{2}}-{1}}}\right|}}}{{{2}\sqrt{{{5}}}}}+{C}$

New Notes Blog

Tuesday, 8 October 2013

$\displaystyle\int\sqrt{{{5}{{x}}^{{2}}-{1}}}{\left.{d}{x}\right.}$ Find the indefinite integral

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Tuesday, 8 October 2013

Find the indefinite integral

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\int\sqrt{{{5}{{x}}^{{2}}-{1}}}{\left.{d}{x}\right.}$ Find the indefinite integral

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?