Question

What is the value of the integral $\int e^{ax}\cos \left(bx\right)dx$?

Answer 1

Find the value of the integral $\int e^{ax}\cos \left(bx\right)dx$.

Let's use integration by parts to solve the above equation.
$\Rightarrow I=\int e^{ax}\cos \left(bx\right)dx$
Let, $u=e^{ax}$ and $v=\cos \left(bx\right)$
$\begin{array}{rcl}\int uvdx& =& u\int vdx-\int u^{\text{'}}\left(\int vdx\right)dx\\ & \Rightarrow & I=1/b{e}^{ax}·\sin bx-a/b\int e^{ax}·\sin bxdx\end{array}$

Apply part integration to $\int e^{ax}\sin bxdx$, we get
$\begin{array}{rcl}I& =& 1/b{e}^{ax}·\sin bx+a/b^2{e}^{ax}·\cos \left(bx\right)-a^2/b^2\int e^{ax}·\cos \left(bx\right)dx\\ & \Rightarrow & I=1/b{e}^{ax}·\sin bx+a/b^2{e}^{ax}·\cos \left(bx\right)-\left(a^2/b^2\right)\mid \end{array}$
Solve both the LHS and the RHS.
$I\left(1+a^2/b^2\right)=1/b{e}^{ax}·\sin bx+a/b^2{e}^{ax}·\cos \left(bx\right)$
By solving the above equations we get
$I=e^{ax}/\left(a^2+b^2\right)\{b\sin bx+a\cos bx\}+c$
Thus, Integral $e^{ax}\cos \left(bx\right)$is $e^{ax}/\left(a^2+b^2\right)\{a\cos bx+b\sin bx\}+c$