Blog: `int e^(2 theta) sin(3 theta) d theta` Evaluate the integral

Thursday, March 1, 2012

$\displaystyle\int{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}{d}\theta$ Evaluate the integral

$\displaystyle\int{{e}}^{{2}}\theta{\sin{{\left({3}\theta\right)}}}{d}\theta$

If f(x) and g(x) are differentiable functions, then

$\displaystyle\int{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}={f{{\left({x}\right)}}}{g{{\left({x}\right)}}}-\int{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}{\left.{d}{x}\right.}$

If we rewrite f(x)=u and g'(x)=v, then

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

Let's integrate using the above method of integration by parts,

Let $\displaystyle{u}={{e}}^{{2}}\theta,{v}={\sin{{\left({3}\theta\right)}}}$

$\displaystyle\int{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}{d}\theta={{e}}^{{{2}\theta}}\int{\sin{{\left({3}\theta\right)}}}{d}\theta-\int{\left(\frac{{d}}{{{d}\theta}}{\left({{e}}^{{{2}\theta}}\right)}\int{\sin{{\left({3}\theta\right)}}}{d}\theta\right)}{d}\theta$

$\displaystyle={{e}}^{{{2}\theta}}{\left(-\frac{{1}}{{3}}{\cos{{\left({3}\theta\right)}}}-\int{\left({{e}}^{{{2}\theta}}{2}{\left(-\frac{{1}}{{3}}{\cos{{\left({3}\theta\right)}}}\right)}{d}\theta\right.}\right.}$

$\displaystyle=-\frac{{1}}{{3}}{{e}}^{{{2}\theta}}{\cos{{\left({3}\theta\right)}}}+\frac{{2}}{{3}}\int{{e}}^{{{2}\theta}}{\cos{{\left({3}\theta\right)}}}{d}\theta$

apply again integration by parts,

$\displaystyle=-\frac{{1}}{{3}}{{e}}^{{{2}\theta}}{\cos{{\left({3}\theta\right)}}}+\frac{{2}}{{3}}{\left({{e}}^{{{2}\theta}}\cdot\int{\cos{{\left({3}\theta\right)}}}{d}\theta-\int{\left({{e}}^{{{2}\theta}}\cdot{2}\int{\cos{{\left({3}\theta\right)}}}{d}\theta\right)}\right.}$

$\displaystyle=-\frac{{1}}{{3}}{{e}}^{{{2}\theta}}{\cos{{\left({3}\theta\right)}}}+\frac{{2}}{{3}}{\left({{e}}^{{{2}\theta}}\frac{{1}}{{3}}{\sin{{\left({3}\theta\right)}}}-\int{2}{{e}}^{{{2}\theta}}{\left(\frac{{\sin{{\left({3}\theta\right)}}}}{{3}}\right)}{d}\theta\right)}$

$\displaystyle=-\frac{{1}}{{3}}{{e}}^{{{2}\theta}}{\cos{{\left({3}\theta\right)}}}+\frac{{2}}{{9}}{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}-\frac{{4}}{{9}}\int{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}{d}\theta$

Isolate $\displaystyle\int{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}{d}\theta$

$\displaystyle{\left({1}+\frac{{4}}{{9}}\right)}\int{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}{d}\theta=-\frac{{1}}{{3}}{{e}}^{{{2}\theta}}{\cos{{\left({3}\theta\right)}}}+\frac{{2}}{{9}}{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}$

$\displaystyle\int{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}{d}\theta=\frac{{9}}{{13}}{\left(-\frac{{1}}{{3}}{{e}}^{{{2}\theta}}{\cos{{\left({3}\theta\right)}}}+\frac{{2}}{{9}}{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}\right)}$

$\displaystyle=-\frac{{3}}{{13}}{{e}}^{{{2}\theta}}{\cos{{\left({3}\theta\right)}}}+\frac{{2}}{{13}}{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}$

add a constant C to the solution,

$\displaystyle=-\frac{{3}}{{13}}{{e}}^{{{2}\theta}}{\cos{{\left({3}\theta\right)}}}+\frac{{2}}{{13}}{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}+{C}$

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Thursday, March 1, 2012

$\displaystyle\int{{e}}^{{{2}\theta}}{\sin{{\left({3}\theta\right)}}}{d}\theta$ Evaluate the integral

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