27. $$ int _ { - 2 } ^ { 3 } e ^ { 2 x } cos 3 x d x $$

Explanation:

Step1: Use integration by parts formula $\int u dv=uv-\int v du$

Let $u = \cos(3x)$, $dv=e^{2x}dx$. Then $du=-3\sin(3x)dx$, $v=\frac{1}{2}e^{2x}$. So, $\int e^{2x}\cos(3x)dx=\frac{1}{2}e^{2x}\cos(3x)+\frac{3}{2}\int e^{2x}\sin(3x)dx$.

Step2: Apply integration by parts again on $\int e^{2x}\sin(3x)dx$

For $\int e^{2x}\sin(3x)dx$, let $u = \sin(3x)$, $dv=e^{2x}dx$. Then $du = 3\cos(3x)dx$, $v=\frac{1}{2}e^{2x}$. So, $\int e^{2x}\sin(3x)dx=\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{2}\int e^{2x}\cos(3x)dx$.

Step3: Substitute back

Substitute $\int e^{2x}\sin(3x)dx$ into the first - integration - by - parts result: [ \begin{align*} \int e^{2x}\cos(3x)dx&=\frac{1}{2}e^{2x}\cos(3x)+\frac{3}{2}\left(\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{2}\int e^{2x}\cos(3x)dx\right)\ \int e^{2x}\cos(3x)dx&=\frac{1}{2}e^{2x}\cos(3x)+\frac{3}{4}e^{2x}\sin(3x)-\frac{9}{4}\int e^{2x}\cos(3x)dx\ \int e^{2x}\cos(3x)dx+\frac{9}{4}\int e^{2x}\cos(3x)dx&=\frac{1}{2}e^{2x}\cos(3x)+\frac{3}{4}e^{2x}\sin(3x)\ \frac{4 + 9}{4}\int e^{2x}\cos(3x)dx&=\frac{1}{2}e^{2x}\cos(3x)+\frac{3}{4}e^{2x}\sin(3x)\ \int e^{2x}\cos(3x)dx&=\frac{e^{2x}(2\cos(3x)+3\sin(3x))}{13}+C \end{align*} ]

Step4: Evaluate the definite integral

[ \begin{align*} \int_{-2}^{3}e^{2x}\cos(3x)dx&=\left[\frac{e^{2x}(2\cos(3x)+3\sin(3x))}{13}\right]_{-2}^{3}\ &=\frac{e^{6}(2\cos(9)+3\sin(9))}{13}-\frac{e^{-4}(2\cos(- 6)+3\sin(-6))}{13}\ &=\frac{e^{6}(2\cos(9)+3\sin(9))}{13}-\frac{e^{-4}(2\cos(6)-3\sin(6))}{13} \end{align*} ]

Answer:

$\frac{e^{6}(2\cos(9)+3\sin(9))}{13}-\frac{e^{-4}(2\cos(6)-3\sin(6))}{13}$