The Notes: `int_0^pi sin^2(t) cos^4(t) dt` Evaluate the integral

Saturday, April 19, 2014

$\displaystyle{\int_{{0}}^{\pi}}{{\sin}}^{{2}}{\left({t}\right)}{{\cos}}^{{4}}{\left({t}\right)}{\left.{d}{t}\right.}$ Evaluate the integral

You need to use the fundamental trigonometric formula $\displaystyle{{\sin}}^{{2}}{x}={1}-{{\cos}}^{{2}}{x}:$

$\displaystyle\int{{\sin}}^{{2}}{t}\cdot{{\cos}}^{{4}}{t}{\left.{d}{t}\right.}=\int{\left({1}-{{\cos}}^{{2}}{t}\right)}\cdot{{\cos}}^{{4}}{t}{\left.{d}{t}\right.}$

$\displaystyle\int{\left({1}-{{\cos}}^{{2}}{t}\right)}\cdot{{\cos}}^{{4}}{t}{\left.{d}{t}\right.}=\int{{\cos}}^{{4}}{t}{\left.{d}{t}\right.}-\int{{\cos}}^{{6}}{t}{\left.{d}{t}\right.}$

You should use the following formula:

$\displaystyle{{\cos}}^{{2}}{t}=\frac{{{1}+{\cos{{2}}}{t}}}{{2}}\Rightarrow{{\cos}}^{{4}}{t}=\frac{{{{\left({1}+{\cos{{2}}}{t}\right)}}^{{2}}}}{{4}}$

$\displaystyle{{\cos}}^{{6}}{t}=\frac{{{{\left({1}+{\cos{{2}}}{t}\right)}}^{{3}}}}{{8}}$

$\displaystyle\int{{\cos}}^{{4}}{t}{\left.{d}{t}\right.}={\left(\frac{{1}}{{4}}\right)}\int{\left({{\left({1}+{\cos{{2}}}{t}\right)}}^{{2}}\right)}{\left.{d}{t}\right.}$

$\displaystyle\int{{\cos}}^{{4}}{t}{\left.{d}{t}\right.}={\left(\frac{{1}}{{4}}\right)}\int{\left.{d}{t}\right.}+{\left(\frac{{1}}{{4}}\right)}\int{2}{\cos{{2}}}{t}{\left.{d}{t}\right.}+{\left(\frac{{1}}{{4}}\right)}{{\cos}}^{{2}}{2}{t}{\left.{d}{t}\right.}$

$\displaystyle\int{{\cos}}^{{4}}{t}{\left.{d}{t}\right.}={\left(\frac{{1}}{{4}}\right)}{t}+{\left(\frac{{1}}{{4}}\right)}{\sin{{2}}}{t}+{\left(\frac{{1}}{{4}}\right)}\int\frac{{{1}+{\cos{{4}}}{t}}}{{2}}{\left.{d}{t}\right.}$

$\displaystyle\int{{\cos}}^{{4}}{t}{\left.{d}{t}\right.}={\left(\frac{{1}}{{4}}\right)}{t}+{\left(\frac{{1}}{{4}}\right)}{\sin{{2}}}{t}+{\left(\frac{{1}}{{8}}\right)}{\left({t}+\frac{{{\sin{{4}}}{t}}}{{4}}\right)}+{c}$

You need to solve $\displaystyle\int{{\cos}}^{{6}}{t}$ dt such that:

$\displaystyle\int{{\cos}}^{{6}}{t}{\left.{d}{t}\right.}=\int\frac{{{{\left({1}+{\cos{{2}}}{t}\right)}}^{{3}}}}{{8}}{\left.{d}{t}\right.}$

$\displaystyle\int{{\cos}}^{{6}}{t}{\left.{d}{t}\right.}={\left(\frac{{1}}{{8}}\right)}\int{\left.{d}{t}\right.}+{\left(\frac{{1}}{{8}}\right)}\int{{\cos}}^{{3}}{2}{t}{\left.{d}{t}\right.}+{\left(\frac{{3}}{{8}}\right)}\int{{\cos}}^{{2}}{2}{t}{\left.{d}{t}\right.}+{\left(\frac{{3}}{{8}}\right)}\int{\cos{{2}}}{t}{\left.{d}{t}\right.}$

$\displaystyle{\left(\frac{{1}}{{8}}\right)}\int{{\cos}}^{{3}}{2}{t}{\left.{d}{t}\right.}={\left(\frac{{1}}{{8}}\right)}\int{{\cos}}^{{2}}{2}{t}\cdot{\cos{{2}}}{t}{\left.{d}{t}\right.}$

$\displaystyle\int{{\cos}}^{{2}}{2}{t}\cdot{\cos{{2}}}{t}{\left.{d}{t}\right.}=\int{\left({1}-{{\sin}}^{{2}}{2}{t}\right)}\cdot{\cos{{2}}}{t}{\left.{d}{t}\right.}$

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

$\displaystyle\int{\left({1}-{{\sin}}^{{2}}{2}{t}\right)}\cdot{\cos{{2}}}{t}{\left.{d}{t}\right.}=\int{\left({1}-{{u}}^{{2}}\right)}\cdot\frac{{{d}{u}}}{{2}}$

$\displaystyle\int{\left({1}-{{u}}^{{2}}\right)}\cdot\frac{{{d}{u}}}{{2}}=\frac{{u}}{{2}}-\frac{{{u}}^{{3}}}{{6}}$

$\displaystyle\int{\left({1}-{{\sin}}^{{2}}{2}{t}\right)}\cdot{\cos{{2}}}{t}{\left.{d}{t}\right.}=\frac{{{\sin{{2}}}{t}}}{{2}}-\frac{{{{\sin}}^{{3}}{2}{t}}}{{6}}$

$\displaystyle\int{{\cos}}^{{6}}{t}{\left.{d}{t}\right.}={\left(\frac{{1}}{{8}}\right)}{t}+{\left(\frac{{1}}{{8}}\right)}{\left(\frac{{{\sin{{2}}}{t}}}{{2}}-\frac{{{{\sin}}^{{3}}{2}{t}}}{{6}}\right)}+{\left(\frac{{3}}{{16}}\right)}{\sin{{2}}}{t}+{\left(\frac{{3}}{{8}}\right)}\int{{\cos}}^{{2}}{2}{t}{\left.{d}{t}\right.}$

Hence, the result of integration is:

$\displaystyle\int{{\sin}}^{{2}}{t}\cdot{{\cos}}^{{4}}{t}{\left.{d}{t}\right.}={\left(\frac{{1}}{{4}}\right)}\pi+{\left(\frac{{1}}{{4}}\right)}{\sin{{2}}}\pi+{\left(\frac{{1}}{{8}}\right)}{\left(\pi+\frac{{{\sin{{8}}}\pi}}{{4}}\right)}-{\left(\frac{{1}}{{8}}\right)}{\left(\pi-{\left(\frac{{1}}{{8}}\right)}{\left(\frac{{{\sin{{2}}}\pi}}{{2}}-\frac{{{{\sin}}^{{3}}{4}\pi}}{{6}}\right)}-{\left(\frac{{3}}{{16}}\right)}{\sin{{4}}}\pi-{\left(\frac{{3}}{{8}}\right)}{\left(\frac{\pi}{{2}}+\frac{{{\sin{{8}}}\pi}}{{4}}\right)}\right.}$

$\displaystyle\int{{\sin}}^{{2}}{t}\cdot{{\cos}}^{{4}}{t}{\left.{d}{t}\right.}=\int{{\sin}}^{{2}}{t}\cdot{{\cos}}^{{4}}{t}{\left.{d}{t}\right.}=\frac{\pi}{{4}}-\frac{{{3}\pi}}{{16}}=\frac{\pi}{{16}}$

The Notes

Saturday, April 19, 2014

$\displaystyle{\int_{{0}}^{\pi}}{{\sin}}^{{2}}{\left({t}\right)}{{\cos}}^{{4}}{\left({t}\right)}{\left.{d}{t}\right.}$ Evaluate the integral

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