The Notes: `int_sqrt(2)^2 1/(t^2 sqrt(t^2 - 1)) dt` Evaluate the integral

Wednesday, July 15, 2015

$\displaystyle{\int_{\sqrt{{{2}}}}^{{2}}}\frac{{1}}{{{{t}}^{{2}}\sqrt{{{{t}}^{{2}}-{1}}}}}{\left.{d}{t}\right.}$ Evaluate the integral

You need to re-write the expression $\displaystyle{{t}}^{{2}}-{1}$ , such that:

$\displaystyle{{t}}^{{2}}-{1}={{t}}^{{2}}{\left({1}-{{\left(\frac{{1}}{{t}}\right)}}^{{2}}\right)}$

You need to use the following substitution, such that:

$\displaystyle\frac{{1}}{{t}}={\sin{{u}}}\Rightarrow-\frac{{1}}{{{{t}}^{{2}}}}{\left.{d}{t}\right.}={\cos{{u}}}{d}{u}\Rightarrow\frac{{{\left.{d}{t}\right.}}}{{{{t}}^{{2}}}}=-{\cos{{u}}}{d}{u}$

$\displaystyle{u}={\arcsin{{\left(\frac{{1}}{{t}}\right)}}}$

$\displaystyle{\int_{{\sqrt{{2}}}}^{{2}}}\frac{{{\left.{d}{t}\right.}}}{{{{t}}^{{3}}\cdot\sqrt{{{{t}}^{{2}}-{1}}}}}=-{\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}\frac{{{\cos{{u}}}{d}{u}}}{{\frac{{1}}{{{\sin{{u}}}}}\cdot\sqrt{{{1}-{{\sin}}^{{2}}{u}}}}}$

You need to use the basic trigonometric formula $\displaystyle{1}-{{\sin}}^{{2}}{u}={{\cos}}^{{2}}{u},$ such that:

$\displaystyle-{\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}\frac{{{\cos{{u}}}\cdot{\sin{{u}}}{d}{u}}}{{\sqrt{{{1}-{{\sin}}^{{2}}{u}}}}}=-{\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}$

$\displaystyle-{\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}\frac{{{\cos{{u}}}\cdot{\sin{{u}}}{d}{u}}}{{\sqrt{{{{\cos}}^{{2}}{u}}}}}=-{\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}\frac{{{\cos{{u}}}\cdot{\sin{{u}}}{d}{u}}}{{{\cos{{u}}}}}$

Reducing like terms yields:

$\displaystyle-{\int_{{{u}_{{1}}}}^{{{u}_{{2}}}}}{\left({\sin{{u}}}{d}{u}\right)}=-{\left(-{\cos{{u}}}\right)}{{\mid}_{{{u}_{{1}}}}^{{{u}_{{2}}}}}$

Replacing back the variable, yields:

$\displaystyle{\int_{{\sqrt{{2}}}}^{{2}}}\frac{{{\left.{d}{t}\right.}}}{{{{t}}^{{3}}\cdot\sqrt{{{{t}}^{{2}}-{1}}}}}={{\left({\cos{{\left({\arcsin{{\left(\frac{{1}}{{t}}\right)}}}\right)}}}\right)}_{{\sqrt{{2}}}}^{{2}}}$

$\displaystyle{\int_{{\sqrt{{2}}}}^{{2}}}\frac{{{\left.{d}{t}\right.}}}{{{{t}}^{{3}}\cdot\sqrt{{{{t}}^{{2}}-{1}}}}}={\cos{{\left({\arcsin{{\left(\frac{{1}}{{2}}\right)}}}\right)}}}-{\cos{{\left({\arcsin{{\left(\frac{{1}}{\sqrt{{2}}}\right)}}}\right)}}}$

$\displaystyle{\int_{{\sqrt{{2}}}}^{{2}}}\frac{{{\left.{d}{t}\right.}}}{{{{t}}^{{3}}\cdot\sqrt{{{{t}}^{{2}}-{1}}}}}={\cos{{\left(\frac{\pi}{{6}}\right)}}}-{\cos{{\left(\frac{\pi}{{4}}\right)}}}$

$\displaystyle{\int_{{\sqrt{{2}}}}^{{2}}}\frac{{{\left.{d}{t}\right.}}}{{{{t}}^{{3}}\cdot\sqrt{{{{t}}^{{2}}-{1}}}}}=\frac{\sqrt{{3}}}{{2}}-\frac{\sqrt{{2}}}{{2}}$

$\displaystyle{\int_{{\sqrt{{2}}}}^{{2}}}\frac{{{\left.{d}{t}\right.}}}{{{{t}}^{{3}}\cdot\sqrt{{{{t}}^{{2}}-{1}}}}}=\frac{{\sqrt{{3}}-\sqrt{{2}}}}{{2}}$

Hence, evaluating the given integral yields $\displaystyle{\int_{{\sqrt{{2}}}}^{{2}}}\frac{{{\left.{d}{t}\right.}}}{{{{t}}^{{3}}\cdot\sqrt{{{{t}}^{{2}}-{1}}}}}=\frac{{\sqrt{{3}}-\sqrt{{2}}}}{{2}}.$

The Notes

Wednesday, July 15, 2015

$\displaystyle{\int_{\sqrt{{{2}}}}^{{2}}}\frac{{1}}{{{{t}}^{{2}}\sqrt{{{{t}}^{{2}}-{1}}}}}{\left.{d}{t}\right.}$ Evaluate the integral

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