`int_0^pi cos^4(2t) dt` Evaluate the integral

Thursday, February 27, 2014

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

Let us call our integral $\displaystyle{I}.$

$\displaystyle{I}={\int_{{0}}^{\pi}}{{\cos}}^{{4}}{2}{t}{\left.{d}{t}\right.}={\int_{{0}}^{\pi}}{{\left({{\cos}}^{{2}}{2}{t}\right)}}^{{2}}{\left.{d}{t}\right.}$

Use formula for cosine of double angle: $\displaystyle{{\cos}}^{{2}}\theta=\frac{{{1}+{\cos{{2}}}\theta}}{{2}}$

$\displaystyle{\int_{{0}}^{\pi}}{{\left(\frac{{{1}+{\cos{{4}}}{t}}}{{2}}\right)}}^{{2}}{\left.{d}{t}\right.}=\frac{{1}}{{4}}{\int_{{0}}^{\pi}}{\left({1}+{2}{\cos{{4}}}{t}+{{\cos}}^{{2}}{4}{t}\right)}{\left.{d}{t}\right.}=$

$\displaystyle\frac{{1}}{{4}}{\int_{{0}}^{\pi}}{\left.{d}{t}\right.}+\frac{{1}}{{2}}{\int_{{0}}^{\pi}}{\cos{{4}}}{t}{\left.{d}{t}\right.}+\frac{{1}}{{4}}{\int_{{0}}^{\pi}}{{\cos}}^{{2}}{4}{t}{\left.{d}{t}\right.}$

Let us denote the above three integrals by $\displaystyle{I}_{{1}},{I}_{{2}}$ and $\displaystyle{I}_{{3}}$ respectively i.e.

$\displaystyle{I}=\frac{{1}}{{4}}{I}_{{1}}+\frac{{1}}{{2}}{I}_{{2}}+\frac{{1}}{{4}}{I}_{{3}}$

$\displaystyle{I}_{{1}}={t}{{\mid}_{{0}}^{\pi}}=\pi$

To solve $\displaystyle{I}_{{2}}$ we make substitution $\displaystyle{u}={4}{t}\Rightarrow\frac{{{d}{u}}}{{4}}={\left.{d}{t}\right.}$ with new limits of integration $\displaystyle{u}_{{1}}={4}\cdot{0}={0}$ and $\displaystyle{u}_{{2}}={4}\cdot\pi={4}\pi.$

$\displaystyle{I}_{{2}}=\frac{{1}}{{4}}{\int_{{0}}^{{{4}\pi}}}{\cos{{u}}}{d}{u}=\frac{{1}}{{4}}{\sin{{u}}}{{\mid}_{{0}}^{{{4}\pi}}}={0}$

To calculate $\displaystyle{I}_{{3}}$ we use formula for cosine of double angle once again.

$\displaystyle{I}_{{3}}={\int_{{0}}^{\pi}}\frac{{{1}+{\cos{{8}}}{t}}}{{2}}{\left.{d}{t}\right.}$

Make substitution $\displaystyle{u}={8}{t}\Rightarrow\frac{{{d}{u}}}{{8}}={\left.{d}{t}\right.}$ with new limits of integration $\displaystyle{u}_{{1}}={8}\cdot{0}={0}$ and $\displaystyle{u}_{{2}}={8}\cdot\pi={8}\pi$

$\displaystyle\frac{{1}}{{16}}{\int_{{0}}^{{{8}\pi}}}{\left({1}+{\cos{{u}}}\right)}{d}{u}=\frac{{1}}{{16}}{\left({u}+{\sin{{u}}}\right)}{{\mid}_{{0}}^{{{8}\pi}}}=\frac{{1}}{{16}}{\left({8}\pi+{0}-{0}-{0}\right)}=\frac{\pi}{{2}}$

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