`int t / sqrt(1-t^4) dt` Find the indefinite integral

Sunday, September 4, 2011

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

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

where: f(x) as the integrand

F(x) as the anti-derivative function

C as the arbitrary constant known as constant of integration

In the given problem: $\displaystyle\int\frac{{t}}{\sqrt{{{1}-{{t}}^{{4}}}}}{\left.{d}{t}\right.}$ , we follow: $\displaystyle\int{f{{\left({t}\right)}}}{\left.{d}{t}\right.}={F}{\left({t}\right)}+{C}.$

The problem can be rewritten as:

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

This resembles the basic integration formula for inverse sine function:

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

Using u-substitution, we let $\displaystyle{u}={{t}}^{{2}}$ then $\displaystyle{d}{u}={2}{t}\cdot{\left.{d}{t}\right.}{\quad\text{or}\quad}\frac{{{d}{u}}}{{2}}={t}\cdot{\left.{d}{t}\right.}$ .

Note: $\displaystyle{{a}}^{{2}}={1}$ then $\displaystyle{a}={1}$

The indefinite integral will be:

$\displaystyle\int\frac{{{t}\cdot{\left.{d}{t}\right.}}}{\sqrt{{{{1}}^{{2}}-{{\left({{t}}^{{2}}\right)}}^{{2}}}}}=\int\frac{{\frac{{{d}{u}}}{{2}}}}{\sqrt{{{{1}}^{{2}}-{{\left({u}\right)}}^{{2}}}}}$

Applying the basic property of integration: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ , we get:

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

Applying the basic integral formula for inverse sine function:

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

$\displaystyle=\frac{{1}}{{2}}{\arcsin{{\left({u}\right)}}}+{C}$

Plug-in $\displaystyle{u}={{t}}^{{2}}$ in $\displaystyle\frac{{1}}{{2}}{\arcsin{{\left({u}\right)}}}+{C}$ to express the indefinite intergral in terms of $\displaystyle\int{f{{\left({t}\right)}}}{\left.{d}{t}\right.}={F}{\left({t}\right)}+{C}$ :

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

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Sunday, September 4, 2011

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

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

Sunday, September 4, 2011

Find the indefinite integral

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

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

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