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

Monday, 2 March 2015

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

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

For easier...

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 easier comparison, we may apply u-substitution by letting $\displaystyle{u}={{e}}^{{x}}$ then $\displaystyle{d}{u}={{e}}^{{x}}{\left.{d}{x}\right.}$ or $\displaystyle\frac{{{d}{u}}}{{{e}}^{{x}}}={\left.{d}{x}\right.}$ .

Note that $\displaystyle{u}={{e}}^{{x}}$ then $\displaystyle\frac{{{d}{u}}}{{{e}}^{{x}}}={\left.{d}{x}\right.}$ becomes $\displaystyle\frac{{{d}{u}}}{{u}}={\left.{d}{x}\right.}$

Plug-in the values on the integral problem, we get:

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

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

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

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

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

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

$\displaystyle\int{{e}}^{{x}}\sqrt{{{1}-{{e}}^{{{2}{x}}}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{{e}}^{{x}}\sqrt{{{1}-{{\left({{e}}^{{x}}\right)}}^{{2}}}}+\frac{{1}}{{2}}{\arctan{{\left(\frac{{{e}}^{{x}}}{\sqrt{{{1}-{{\left({{e}}^{{x}}\right)}}^{{2}}}}}\right)}}}+{C}$

$\displaystyle=\frac{{1}}{{2}}{{e}}^{{x}}\sqrt{{{1}-{{e}}^{{{2}{x}}}}}+\frac{{1}}{{2}}{\arctan{{\left(\frac{{{e}}^{{x}}}{\sqrt{{{1}-{{e}}^{{{2}{x}}}}}}\right)}}}+{C}$

$\displaystyle=\frac{{{{e}}^{{x}}\sqrt{{{1}-{{e}}^{{{2}{x}}}}}}}{{2}}+\frac{{\arctan{{\left(\frac{{{e}}^{{x}}}{\sqrt{{{1}-{{e}}^{{{2}{x}}}}}}\right)}}}}{{2}}+{C}$

New Notes Blog

Monday, 2 March 2015

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

No comments:

Post a Comment

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

Monday, 2 March 2015

Find the indefinite integral

No comments:

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?

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