The Notes: `int (dt)/sqrt(t^2 - 6t + 13)` Evaluate the integral

Tuesday, January 27, 2009

$\displaystyle\int\frac{{{\left.{d}{t}\right.}}}{\sqrt{{{{t}}^{{2}}-{6}{t}+{13}}}}$ Evaluate the integral

$\displaystyle\int\frac{{\left.{d}{t}}}{\sqrt{{{{t}}^{{2}}-{6}{t}+{13}}}}$

Let's evaluate the integral by rewriting it by completing the square on the denominator,

$\displaystyle=\int\frac{{\left.{d}{t}}}{\sqrt{{{{\left({t}-{3}\right)}}^{{2}}+{4}}}}$

Now let's use the integral substitution,

Let $\displaystyle{u}={t}-{3}$

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

$\displaystyle=\int\frac{{{d}{u}}}{\sqrt{{{{u}}^{{2}}+{4}}}}$

Now use the trigonometric substitution: For $\displaystyle\sqrt{{{b}{{x}}^{{2}}+{a}}}$ substitute $\displaystyle{x}=\frac{\sqrt{{{a}}}}{\sqrt{{{b}}}}{\tan{{\left({v}\right)}}}$

So ,Let $\displaystyle{u}={2}{\tan{{\left({v}\right)}}}$

$\displaystyle\Rightarrow{d}{u}={2}{{\sec}}^{{2}}{\left({v}\right)}{d}{v}$

$\displaystyle=\int\frac{{{2}{{\sec}}^{{2}}{\left({v}\right)}{d}{v}}}{\sqrt{{{{\left({2}{\tan{{\left({v}\right)}}}\right)}}^{{2}}+{4}}}}$

$\displaystyle=\int\frac{{{2}{{\sec}}^{{2}}{\left({v}\right)}{d}{v}}}{\sqrt{{{4}{{\tan}}^{{2}}{\left({v}\right)}+{4}}}}$

$\displaystyle=\int\frac{{{2}{{\sec}}^{{2}}{\left({v}\right)}{d}{v}}}{\sqrt{{{4}{\left({{\tan}}^{{2}}{\left({v}\right)}+{1}\right)}}}}$

$\displaystyle=\int\frac{{{2}{{\sec}}^{{2}}{\left({v}\right)}{d}{v}}}{{{2}\sqrt{{{{\tan}}^{{2}}{\left({v}\right)}+{1}}}}}$

$\displaystyle=\int\frac{{{{\sec}}^{{2}}{\left({v}\right)}{d}{v}}}{\sqrt{{{{\tan}}^{{2}}{\left({v}\right)}+{1}}}}$

Now use the identity: $\displaystyle{1}+{{\tan}}^{{2}}{\left({x}\right)}={{\sec}}^{{2}}{\left({x}\right)}$

$\displaystyle=\int\frac{{{{\sec}}^{{2}}{\left({v}\right)}{d}{v}}}{\sqrt{{{{\sec}}^{{2}}{\left({v}\right)}}}}$

$\displaystyle=\int{\sec{{\left({v}\right)}}}{d}{v}$

Now use the standard integral.

$\displaystyle\int{\sec{{\left({x}\right)}}}{\left.{d}{x}\right.}={\ln}{\left|{\sec{{\left({x}\right)}}}+{\tan{{\left({x}\right)}}}\right|}$

$\displaystyle={\ln}{\left|{\sec{{\left({v}\right)}}}+{\tan{{\left({v}\right)}}}\right|}$

Substitute back $\displaystyle{\tan{{\left({v}\right)}}}=\frac{{u}}{{2}}$

$\displaystyle\Rightarrow{1}+{{\tan}}^{{2}}{\left({v}\right)}={{\sec}}^{{2}}{\left({v}\right)}$

$\displaystyle\Rightarrow{1}+{{\left(\frac{{u}}{{2}}\right)}}^{{2}}={{\sec}}^{{2}}{\left({v}\right)}$

$\displaystyle\Rightarrow{{\sec}}^{{2}}{\left({v}\right)}=\frac{{{{u}}^{{2}}+{4}}}{{4}}$

$\displaystyle\Rightarrow{\sec{{\left({v}\right)}}}=\frac{\sqrt{{{{u}}^{{2}}+{4}}}}{{2}}$

Plug these into the solution, thus

$\displaystyle={\ln}{\left|\frac{\sqrt{{{{u}}^{{2}}+{4}}}}{{2}}+\frac{{u}}{{2}}\right|}$

Now plug back u=t-3 and add a constant C to the solution,

$\displaystyle={\ln}{\left|\frac{\sqrt{{{{\left({t}-{3}\right)}}^{{2}}+{4}}}}{{2}}+\frac{{{t}-{3}}}{{2}}\right|}+{C}$

The Notes

Tuesday, January 27, 2009

$\displaystyle\int\frac{{{\left.{d}{t}\right.}}}{\sqrt{{{{t}}^{{2}}-{6}{t}+{13}}}}$ Evaluate the integral

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