Blog: `int_(1/6)^(1/2) csc(pi t) cot(pi t) dt` Evaluate the definite integral.

Saturday, August 24, 2013

Given $\displaystyle{\int_{{\frac{{1}}{{6}}}}^{{\frac{{1}}{{2}}}}}{\csc{{\left(\pi{t}\right)}}}{\cot{{\left(\pi{t}\right)}}}{\left.{d}{t}\right.}$

Integrate using the Substitution Rule.

Let $\displaystyle{u}=\pi{t}$

$\displaystyle\frac{{{d}{u}}}{{\left.{d}{t}}}=\pi$

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

$\displaystyle={\int_{{\frac{{1}}{{6}}}}^{{\frac{{1}}{{2}}}}}{\csc{{\left({u}\right)}}}{\cot{{\left({u}\right)}}}\cdot\frac{{{d}{u}}}{\pi}$

$\displaystyle=\frac{{1}}{\pi}{\int_{{\frac{{1}}{{6}}}}^{{\frac{{1}}{{2}}}}}{\csc{{\left({u}\right)}}}{\cot{{\left({u}\right)}}}{d}{u}$

$\displaystyle=\frac{{1}}{\pi}\cdot{\left[-{\csc{{\left({u}\right)}}}\right]}$ Evaluated from t=1/6 to t=1/2

$\displaystyle=-\frac{{1}}{\pi}{\csc{{\left({u}\right)}}}$ Evaluated from t=1/6 to t=1/2

Right now the limits of integration are in terms of t. Change the limits of integration to terms of u.

Since $\displaystyle{u}=\pi{t}$

When $\displaystyle{t}=\frac{{1}}{{6}}$ , $\displaystyle{u}=\frac{\pi}{{6}}$

When $\displaystyle{t}=\frac{{1}}{{2}}$ , $\displaystyle{u}=\frac{\pi}{{2}}$

$\displaystyle=-\frac{{1}}{\pi}{\csc{{\left({u}\right)}}}$ Evaluated from $\displaystyle{u}=\frac{\pi}{{6}}$ to $\displaystyle{u}=\frac{\pi}{{2}}$

$\displaystyle=\frac{{1}}{{-\pi}}{\left[{\csc{{\left(\frac{\pi}{{2}}\right)}}}-{\csc{{\left(\frac{\pi}{{6}}\right)}}}\right]}$

$\displaystyle=\frac{{1}}{{-{\pi}}}{\left[{1}-{2}\right]}$

$\displaystyle=\frac{{1}}{{-{\pi}}}{\left[-{1}\right]}$

$\displaystyle=\frac{{1}}{\pi}$

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