Blog: `int e^(-theta) cos(2 theta) d theta` Evaluate the integral

Monday, August 29, 2016

$\displaystyle\int{{e}}^{{-\theta}}{\cos{{\left({2}\theta\right)}}}{d}\theta$ Evaluate the integral

2∫(-2excos(2x))dx+2exsin(2x)-1excos(2x)=2(-2∫1excos(2x)dx)+2exsin(2x)-1excos(2x)

Use integration by parts:

$\displaystyle\int{\left({u}\right)}{d}{v}={u}{v}-\int{v}{d}{u}$

$\displaystyle{L}{e}{t}{u}={\cos{{\left({2}{x}\right)}}}{\quad\text{and}\quad}{d}{v}=\frac{{\left.{d}{x}}}{{{e}}^{{x}}}$

Then,

$\displaystyle{d}{u}=-{2}{\sin{{2}}}{x}{\left.{d}{x}\right.}$

$\displaystyle{v}=\int{\left({{e}}^{{-{{x}}}}\right)}{\left.{d}{x}\right.}=-\frac{{1}}{{{e}}^{{x}}}$

Thus,

$\displaystyle\int{\left(\frac{{1}}{{{e}}^{{x}}}{\cos{{\left({2}{x}\right)}}}\right)}{\left.{d}{x}\right.}={\left({\cos{{\left({2}{x}\right)}}}⋅{\left(-\frac{{1}}{{{e}}^{{x}}}\right)}−\int{\left({\left(-\frac{{1}}{{{e}}^{{x}}}\right)}⋅{\left(-{2}{\sin{{\left({2}{x}\right)}}}\right)}{\left.{d}{x}\right.}\right)}\right)}={\left(-\int{\left(\frac{{2}}{{{e}}^{{x}}}{\sin{{\left({2}{x}\right)}}}{\left.{d}{x}\right.}-\frac{{1}}{{{e}}^{{x}}}{\cos{{\left({2}{x}\right)}}}\right)}\right)}$

Use the constant multiple rule

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

Use integration by parts again for the first part

Let $\displaystyle{u}={\sin{{\left({2}{x}\right)}}}{\quad\text{and}\quad}{d}{v}=\frac{{\left.{d}{x}}}{{{e}}^{{x}}}$

Then,

$\displaystyle{d}{u}={2}{\cos{{\left({2}{x}\right)}}}{\left.{d}{x}\right.}{\quad\text{and}\quad}{v}=\int{\left({{e}}^{{-{{x}}}}\right)}{\left.{d}{x}\right.}=-\frac{{1}}{{{e}}^{{-{{x}}}}}$

Integral becomes

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

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

Apply constant multiple rule

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

Simplify

$\displaystyle\int{\left(\frac{{1}}{{{e}}^{{x}}}{\cos{{\left({2}{x}\right)}}}\right)}{\left.{d}{x}\right.}=\frac{{1}}{{{5}{{e}}^{{x}}}}{\left({2}{\sin{{\left({2}{x}\right)}}}-{\cos{{\left({2}{x}\right)}}}\right)}+{C}$

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Monday, August 29, 2016

$\displaystyle\int{{e}}^{{-\theta}}{\cos{{\left({2}\theta\right)}}}{d}\theta$ Evaluate the integral

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