The Notes: `int sqrt(5 + 4x - x^2) dx` Evaluate the integral

Sunday, June 19, 2016

$\displaystyle\int\sqrt{{{5}+{4}{x}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int\sqrt{{{5}+{4}{x}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$

Rewrite the integrand by completing the square:

$\displaystyle=\int\sqrt{{-{{\left({x}-{2}\right)}}^{{2}}+{9}}}{\left.{d}{x}\right.}$

Now apply the integral substitution,

Let u=x-2,

$\displaystyle\Rightarrow{d}{u}={\left.{d}{x}\right.}$

$\displaystyle=\int\sqrt{{{9}-{{u}}^{{2}}}}{d}{u}$

Now using the standard integral:

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

$\displaystyle\int\sqrt{{{9}-{{u}}^{{2}}}}{d}{u}=\frac{{{u}\sqrt{{{{3}}^{{2}}-{{u}}^{{2}}}}}}{{2}}+\frac{{{3}}^{{2}}}{{2}}{{\sin}}^{{-{1}}}\frac{{u}}{{3}}$

$\displaystyle=\frac{{{u}\sqrt{{{9}-{{u}}^{{2}}}}}}{{2}}+\frac{{9}}{{2}}{{\sin}}^{{-{1}}}\frac{{u}}{{3}}$

Substitute back u=x-2,

$\displaystyle=\frac{{{\left({x}-{2}\right)}\sqrt{{{9}-{{\left({x}-{2}\right)}}^{{2}}}}}}{{2}}+\frac{{9}}{{2}}{{\sin}}^{{-{1}}}{\left(\frac{{{x}-{2}}}{{3}}\right)}$

Add a constant C to the solution,

$\displaystyle=\frac{{{\left({x}-{2}\right)}\sqrt{{{9}-{{\left({x}-{2}\right)}}^{{2}}}}}}{{2}}+\frac{{9}}{{2}}{{\sin}}^{{-{1}}}{\left(\frac{{{x}-{2}}}{{3}}\right)}+{C}$

The Notes

Sunday, June 19, 2016

$\displaystyle\int\sqrt{{{5}+{4}{x}-{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Evaluate the integral

No comments:

Post a Comment

What are hearing tests?