The Notes: `int sqrt(1 - 4x^2) dx` Evaluate the integral

$\displaystyle\int\sqrt{{{1}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}$

$\displaystyle=\int\sqrt{{{1}-{{\left({2}{x}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

Now apply the integral substitution,

Let u=2x,

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

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

Now using the standard integral,

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

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

Now substitute back u=2x,

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

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

The Notes

Sunday, December 2, 2012

$\displaystyle\int\sqrt{{{1}-{4}{{x}}^{{2}}}}{\left.{d}{x}\right.}$ Evaluate the integral

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