Question

Evaluate the integral displaystyleint{e}^{{{3}{x}}} cos{{2}}{x}{left.{d}{x}right.}

Integrals
ANSWERED
asked 2021-03-09
Evaluate the integral \(\displaystyle\int{e}^{{{3}{x}}} \cos{{2}}{x}{\left.{d}{x}\right.}\)

Answers (1)

2021-03-10
You just need to write your expression in terms of only exponentials, do the resulting very easy integral, and then do a bunch of algebra to get it back into a form where you can see the \(\displaystyle \sin{'}{s}{\quad\text{and}\quad} \cos{'}{s}\).
\(\displaystyle\int{e}^{{{3}{x}}} \cos{{\left({2}{x}\right)}}{\left.{d}{x}\right.}=\int{e}^{{{3}{x}}}\frac{1}{{2}}{\left[{e}^{{{2}{x}{i}}}+{e}^{{-{2}{x}{i}}}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle=\frac{1}{{2}}{\left[{e}^{{{\left({3}+{2}{i}\right)}{x}}}+{e}^{{{\left({3}-{2}{i}\right)}{x}}}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle=\frac{1}{{2}}{\left[\frac{{{e}^{{{\left({3}+{2}{i}\right)}}}{x}}}{{{3}+{2}{i}}}+\frac{{{e}^{{{\left({3}-{2}{i}\right)}}}{x}}}{{{3}-{2}{i}}}\right]}\)
\(\displaystyle=\frac{1}{{2}}{\left[\frac{{{\left({3}-{2}{i}\right)}{e}^{{{\left({3}+{2}{i}\right)}}}{x}+{\left({3}+{2}{i}\right)}{e}^{{{\left({3}-{2}\right)}{i}}}{x}}}{{13}}\right]}\)
\(\displaystyle=\frac{{{e}^{{{3}{x}}}}}{{26}}{\left[{\left({3}-{2}{i}\right)}{e}^{{{2}{x}{i}}}+{\left({3}+{2}{i}\right)}{e}^{{-{2}{x}{i}}}\right]}\)
\(\displaystyle=\frac{{e}^{{{3}{x}}}}{{13}}{\left[{3}\cdot\frac{{{e}^{{{2}{x}{i}}}+{e}^{{-{2}{x}{i}}}}}{{2}}+{2}\cdot\frac{{{e}^{{{2}{x}{i}}}-{e}^{{-{2}{x}{i}}}}}{{{2}{i}}}\right]}\)
\(\displaystyle\frac{{e}^{{{3}{x}}}}{{13}}{\left[{3} \cos{{\left({2}{x}\right)}}+{2} \sin{{\left({2}{x}\right)}}\right]}\)
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