Question

$$\int e^{ax}\text{ sin}(bx+C)dx$$

Answer 1

$$\text{ Let I }=\int e^{ax}\sin (bx+C)\text{ dx }$$
$$\text{ Considering sin}(bx+C)as\text{ first function and }{e}^{ax}\text{ as second function}$$
$$I=\text{ sin }(bx+C)\frac{e^{ax}}{a}-\int \text{ cos }(bx+C)b\frac{e^{ax}}{a}dx$$
$$\Rightarrow I=\frac{e^{ax}\text{ sin }(bx+C)}{a}-\frac{b}{a}\int e^{ax}\text{ cos}(bx+C)dx$$
$$\Rightarrow I=\frac{e^{ax}\text{ sin }(bx+C)}{a}-\frac{b}{a}{I}_1...\left(1\right)$$
$${\text{ where I}}_1=\int e^{ax}\text{ cos }(bx+C)dx$$
$$\text{ Now, }{I}_1=\int e^{ax}\cos (bx+C)dx$$
$$\text{ Consider cos}(bx+C)\text{ as first function }{e}^{ax}\text{ as second funciton }$$
$$I_1=\cos (bx+C)\frac{e^{ax}}{a}-\int -\sin (bx+C)b\frac{e^{ax}}{a}dx$$
$$\Rightarrow I_1=\frac{e^{ax}\cos (bx+C)}{a}+\frac{b}{a}\int e^{ax}\sin (bx+C)dx$$
$$\Rightarrow I_1=\frac{e^{ax}\cos (bx+C)}{a}+\frac{b}{a}I.....\left(2\right)$$
${\displaystyle \text{From ( 1 ) and ( 2 )}}$
$$I=\frac{e^{ax}\sin (bx+C)}{a}-\frac{b}{a}[\frac{e^{ax}\cos (bx+C)}{a}+\frac{b}{a}I]$$
$$\Rightarrow I=\frac{e^{ax}\sin (bx+C)}{a}-\frac{b}{a^2}{e}^{ax}\text{ cos }(bx+C)-\frac{b^2}{a^2}I$$
$$\Rightarrow I(1+\frac{b^2}{a^2})=\frac{e^{ax}\text{ a }\text{ sin }(bx+C)-b{e}^{ax}\text{ cos }(bx+C)}{a^2}+C_1$$
$$\Rightarrow I=e^{ax}\frac{[a\text{ sin }(bx+C)-b\text{ cos}(bx+C)]}{a^2+b^2}+C_1$$
${\displaystyle {\text{Where C}}_1}$  ${\displaystyle \text{is integration constant}}$