New Notes Blog: `int xarcsec(x^2+1) dx` Use integration tables to find the indefinite integral.

Sunday, 8 December 2013

$\displaystyle\int{x}{a}{r}{c}{\sec{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}$ Use integration tables to find the indefinite integral.

Indefinite integral are written in the form of $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

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 $\displaystyle\int{x}{a}{r}{c}{\sec{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.},$ it has a integrand in a form of inverse secant function. The integral resembles one of the formulas from the integration as : $\displaystyle\int{a}{r}{c}{\sec{{\left(\frac{{u}}{{a}}\right)}}}{d}{u}={u}\cdot{\arcsin{{\left(\frac{{u}}{{a}}\right)}}}\pm{a}{\ln{{\left({u}+\sqrt{{{{u}}^{{2}}-{{a}}^{{2}}}}\right)}}}+{C}$ .

where we use: $\displaystyle{\left(+\right)}$ if $\displaystyle{0}\lt{a}{r}{c}{\sec{{\left(\frac{{u}}{{a}}\right)}}}\lt\frac{\pi}{{2}}$

$\displaystyle{\left(-\right)}$ if $\displaystyle\frac{\pi}{{2}}\lt{a}{r}{c}{\sec{{\left(\frac{{u}}{{a}}\right)}}}\lt\pi$

Selecting the sign between $\displaystyle{\left(+\right)}$ and $\displaystyle{\left(-\right)}$ will be crucial when solving for definite integral with given boundary values $\displaystyle{\left[{a},{b}\right]}$ .

For easier comparison, we may apply u-substitution by letting:

$\displaystyle{u}={{x}}^{{2}}+{1}$ then $\displaystyle{d}{u}={2}{x}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{2}}$

Plug-in the values $\displaystyle\int{x}{a}{r}{c}{\sec{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}$ , we get:

$\displaystyle\int{x}{a}{r}{c}{\sec{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}=\int{a}{r}{c}{\sec{{\left({{x}}^{{2}}+{1}\right)}}}\cdot{x}{\left.{d}{x}\right.}$

$\displaystyle=\int{a}{r}{c}{\sec{{\left({u}\right)}}}\cdot\frac{{{d}{u}}}{{2}}$

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

$\displaystyle\int{a}{r}{c}{\sec{{\left({u}\right)}}}\cdot\frac{{{d}{u}}}{{2}}=\frac{{1}}{{2}}\int{a}{r}{c}{\sec{{\left({u}\right)}}}{d}{u}$

or $\displaystyle\frac{{1}}{{2}}\int{a}{r}{c}{\sec{{\left(\frac{{u}}{{1}}\right)}}}{d}{u}$

Applying the aforementioned formula from the integration table, we get:

$\displaystyle\frac{{1}}{{2}}\int{a}{r}{c}{\sec{{\left(\frac{{u}}{{1}}\right)}}}{d}{u}=\frac{{1}}{{2}}\cdot{\left[{u}\cdot{\arcsin{{\left(\frac{{u}}{{1}}\right)}}}\pm{1}{\ln{{\left({u}+\sqrt{{{{u}}^{{2}}-{{1}}^{{2}}}}\right)}}}\right]}+{C}$

$\displaystyle=\frac{{1}}{{2}}\cdot{\left[{u}\cdot{\arcsin{{\left({u}\right)}}}\pm{\ln{{\left({u}+\sqrt{{{{u}}^{{2}}-{1}}}\right)}}}\right]}+{C}$

$\displaystyle=\frac{{{u}\cdot{\arcsin{{\left({u}\right)}}}}}{{2}}\pm\frac{{{\ln{{\left({u}+\sqrt{{{{u}}^{{2}}-{1}}}\right)}}}}}{{2}}+{C}$

Plug-in $\displaystyle{u}={{x}}^{{2}}+{1}$ on $\displaystyle\frac{{{u}\cdot{\arcsin{{\left({u}\right)}}}}}{{2}}\pm\frac{{{\ln{{\left({u}+\sqrt{{{{u}}^{{2}}-{1}}}\right)}}}}}{{2}}+{C}$ , we get the indefinite integral as:

$\displaystyle\int{x}{a}{r}{c}{\sec{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}=\frac{{{\left({{x}}^{{2}}+{1}\right)}\cdot{\arcsin{{\left({{x}}^{{2}}+{1}\right)}}}}}{{2}}\pm\frac{{{\ln{{\left({{x}}^{{2}}+{1}+\sqrt{{{{\left({{x}}^{{2}}+{1}\right)}}^{{2}}-{1}}}\right)}}}}}{{2}}+{C}$

$\displaystyle=\frac{{{{x}}^{{2}}{\arcsin{{\left({{x}}^{{2}}+{1}\right)}}}}}{{2}}+\frac{{\arcsin{{\left({{x}}^{{2}}+{1}\right)}}}}{{2}}\pm\frac{{\ln{{\left({{x}}^{{2}}+{1}+\sqrt{{{{x}}^{{4}}+{2}{{x}}^{{2}}}}\right)}}}}{{2}}+{C}$

$\displaystyle=\frac{{{{x}}^{{2}}{\arcsin{{\left({{x}}^{{2}}+{1}\right)}}}}}{{2}}+\frac{{\arcsin{{\left({{x}}^{{2}}+{1}\right)}}}}{{2}}\pm\frac{{\ln{{\left({{x}}^{{2}}+{1}+{\left|{x}\right|}\sqrt{{{{x}}^{{2}}+{2}}}\right)}}}}{{2}}+{C}$

New Notes Blog

Sunday, 8 December 2013

$\displaystyle\int{x}{a}{r}{c}{\sec{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}$ Use integration tables to 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?

Sunday, 8 December 2013

Use integration tables to 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?

$\displaystyle\int{x}{a}{r}{c}{\sec{{\left({{x}}^{{2}}+{1}\right)}}}{\left.{d}{x}\right.}$ Use integration tables to 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?