`int (dt)/(t^2 sqrt(t^2 - 16))` Evaluate the integral

Wednesday, October 12, 2011

$\displaystyle\int\frac{{{\left.{d}{t}\right.}}}{{{{t}}^{{2}}\sqrt{{{{t}}^{{2}}-{16}}}}}$ Evaluate the integral

$\displaystyle\int\frac{{1}}{{{{t}}^{{2}}\sqrt{{{{t}}^{{2}}-{16}}}}}{\left.{d}{t}\right.}$

Let's evaluate the integral by applying integral substitution,

Let $\displaystyle{t}={4}{\sec{{\left({u}\right)}}}$

$\displaystyle\Rightarrow{\left.{d}{t}\right.}={4}{\tan{{\left({u}\right)}}}{\sec{{\left({u}\right)}}}{d}{u}$

Plug the above in the integral,

$\displaystyle\int\frac{{1}}{{{{t}}^{{2}}\sqrt{{{{t}}^{{2}}-{16}}}}}{\left.{d}{t}\right.}=\int\frac{{{4}{\tan{{\left({u}\right)}}}{\sec{{\left({u}\right)}}}}}{{{\left({{4}}^{{2}}{{\sec}}^{{2}}{\left({u}\right)}\sqrt{{{{4}}^{{2}}{{\sec}}^{{2}}{\left({u}\right)}-{16}}}\right)}}}{d}{u}$

$\displaystyle=\int\frac{{{4}{\tan{{\left({u}\right)}}}{\sec{{\left({u}\right)}}}}}{{{16}{{\sec}}^{{2}}{\left({u}\right)}\cdot{4}\sqrt{{{{\sec}}^{{2}}{\left({u}\right)}-{1}}}}}{d}{u}$

$\displaystyle=\int\frac{{\tan{{\left({u}\right)}}}}{{{16}{\sec{{\left({u}\right)}}}\sqrt{{{{\sec}}^{{2}}{\left({u}\right)}-{1}}}}}{d}{u}$

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

$\displaystyle=\frac{{1}}{{16}}\int\frac{{\tan{{\left({u}\right)}}}}{{{\sec{{\left({u}\right)}}}\sqrt{{{1}+{{\tan}}^{{2}}{\left({u}\right)}-{1}}}}}{d}{u}$

$\displaystyle=\frac{{1}}{{16}}\int\frac{{\tan{{\left({u}\right)}}}}{{{\sec{{\left({u}\right)}}}{\tan{{\left({u}\right)}}}}}{d}{u}$

$\displaystyle=\frac{{1}}{{16}}\int\frac{{1}}{{\sec{{\left({u}\right)}}}}{d}{u}$

$\displaystyle=\frac{{1}}{{16}}\int{\cos{{\left({u}\right)}}}{d}{u}$

Now use the standard integral: $\displaystyle\int{\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}={\sin{{\left({x}\right)}}}+{C}$

$\displaystyle=\frac{{1}}{{16}}{\sin{{\left({u}\right)}}}+{C}$

We have used $\displaystyle{t}={4}{\sec{{\left({u}\right)}}}$

$\displaystyle\Rightarrow{u}={a}{r}{c}{\sec{{\left(\frac{{t}}{{4}}\right)}}}$

Substitute back u in the solution,

$\displaystyle=\frac{{1}}{{16}}{\sin{{\left({a}{r}{c}{\sec{{\left(\frac{{t}}{{4}}\right)}}}\right)}}}+{C}$

Simplify the above by assuming the right triangle with angle $\displaystyle\theta={a}{r}{c}{\sec{{\left(\frac{{t}}{{4}}\right)}}}$

Hypotenuse is t and adjacent side is 4, Opposite side(O) can be found by using pythagorean identity,

$\displaystyle{{4}}^{{2}}+{{O}}^{{2}}={{t}}^{{2}}$

$\displaystyle{{O}}^{{2}}={{t}}^{{2}}-{16}$

$\displaystyle{O}=\sqrt{{{{t}}^{{2}}-{16}}}$

$\displaystyle{\sin{{\left(\theta\right)}}}=\frac{\sqrt{{{{t}}^{{2}}-{16}}}}{{t}}$

Hence the solution is $\displaystyle\frac{{1}}{{16}}{\left(\frac{\sqrt{{{{t}}^{{2}}-{16}}}}{{t}}\right)}+{C}$

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Wednesday, October 12, 2011

$\displaystyle\int\frac{{{\left.{d}{t}\right.}}}{{{{t}}^{{2}}\sqrt{{{{t}}^{{2}}-{16}}}}}$ Evaluate the 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?

Wednesday, October 12, 2011

Evaluate the 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?

$\displaystyle\int\frac{{{\left.{d}{t}\right.}}}{{{{t}}^{{2}}\sqrt{{{{t}}^{{2}}-{16}}}}}$ Evaluate the integral

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