Question

Write a value of $\int e^{ax}\cos bxdx$.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int e^{ax}.\cos bxdx \\ =\cos bx\int e^{ax}dx-\int \left\{\frac{d}{dx}\left(\cos bx\right)\int e^{ax}dx\right\}dx \\ =\cos bx\times \frac{e^{ax}}{a}-\int -\sin bx\times b.\frac{e^{ax}}{a} \\ =\cos bx\times \frac{e^{ax}}{a}+\frac{b}{a}\int e^{ax}.\sin bxdx \\ =\cos bx\times \frac{e^{ax}}{a}+\frac{b}{a}{I}_1...\left(1\right) \\ \therefore I_1=\int e^{ax}\times \sin bxdx \\ =\sin bx\int e^{ax}dx-\int \left\{\frac{d}{dx}\left(\sin bx\right)\int e^{ax}dx\right\}dx \\ =\sin bx\times \frac{e^{ax}}{a}-\int b.\cos bx\times \frac{e^{ax}}{a}dx \\ =\sin bx.\frac{e^{ax}}{a}-\frac{b}{a}I\mathit{}....\left(2\right) \\ \mathrm{From}\left(1\right)\&\left(2\right) \\ \therefore I=\cos bx.\frac{e^{ax}}{a}+\frac{b}{a}\left\{\sin bx.\frac{e^{ax}}{a}-\frac{b}{a}I\right\} \\ \Rightarrow I=\cos bx.\frac{e^{ax}}{a}+\frac{b}{a^2}\sin bx{e}^{ax}-\frac{b^2}{a^2}I \\ \Rightarrow I+\frac{b^2}{a^2}I=\cos bx.\frac{e^{ax}}{a}+\frac{b\sin bx{e}^{ax}}{a^2} \\ \Rightarrow \left(a^2+b^2\right)I=\left(a\cos bx+b\sin bx\right)e^{ax} \\ \Rightarrow I=\frac{\left(a\cos bx+b\sin bx\right)e^{ax}}{a^2+b^2}+C \end{gathered}$$