Question

Find value of $\int {{e^{ax}} \cdot \sin \left( {bx + c} \right)dx} $

Answer 1

Hint:
let us consider $I = \int {{e^{ax}} \cdot \sin \left( {bx + c} \right)dx} $ then we will use integration by part $\int {UVdx = U\int {Vdx - \int {\dfrac{{dU}}{{dx}}\left( {\int {Vdx} } \right)dx} } } $ where $U = \sin \left( {bx + c} \right)$ and $V = {e^{ax}}$. And after that we can see that I will be in terms of cosine then we will use integration by part again and convert into sine form.

Complete step by step solution:
Let $I = \int {{e^{ax}} \cdot \sin \left( {bx + c} \right)dx} $
Now, using integration by part $\int {UVdx = U\int {Vdx - \int {\dfrac{{dU}}{{dx}}\left( {\int {Vdx} } \right)dx} } } $ where $U = \sin \left( {bx + c} \right)$ and $V = {e^{ax}}$
So, $I = \sin \left( {bx + c} \right)\int {{e^{ax}}dx - \int {\dfrac{{d\sin \left( {bx + c} \right)}}{{dx}}\left( {\int {{e^{ax}}dx} } \right)dx} } $
$\int {{e^{ax}}dx = \dfrac{{{e^{ax}}}}{a}} $ and $\dfrac{{d\sin \left( {bx + c} \right)}}{{dx}} = b\cos \left( {bx + c} \right)$
\[I = \dfrac{{\sin \left( {bx + c} \right){e^{ax}} - \int {b\cos \left( {bx + c} \right){e^{ax}}dx} }}{a}\]
Again, using integration by part for \[\int {b\cos \left( {bx + c} \right){e^{ax}}dx} \] where $U = \cos \left( {bx + c} \right)$ and $V = {e^{ax}}$
\[I = \dfrac{1}{a}\left( {\sin \left( {bx + c} \right){e^{ax}} - \left( {b\cos \left( {bx + c} \right)\int {{e^{ax}}dx - b\int {\dfrac{{d\cos \left( {bx + c} \right)}}{{dx}}\left( {\int {{e^{ax}}} dx} \right)dx} } } \right)} \right)\]
$\int {{e^{ax}}dx = \dfrac{{{e^{ax}}}}{a}} $ and $\dfrac{{d\cos \left( {bx + c} \right)}}{{dx}} = - b\sin \left( {bx + c} \right)$
\[I = \dfrac{1}{a}\left( {\sin \left( {bx + c} \right){e^{ax}} - \left( {\dfrac{b}{a}\cos \left( {bx + c} \right){e^{ax}} - b\int { - \dfrac{b}{a}\sin \left( {bx + a} \right){e^{ax}}dx} } \right)} \right)\]
$I = \int {{e^{ax}} \cdot \sin \left( {bx + c} \right)dx} $
So, \[I = \dfrac{1}{a}\sin \left( {bx + c} \right){e^{ax}} - \left( {\dfrac{b}{{{a^2}}}\cos \left( {bx + c} \right){e^{ax}} + \left( {\dfrac{{{b^2}}}{{{a^2}}}I} \right)} \right)\]
$\dfrac{{{a^2} - {b^2}}}{{{a^2}}}I = \dfrac{{{e^{ax}}}}{{{a^2}}}\left( {\sin \left( {bx + c} \right) - b\cos \left( {bx + c} \right)} \right)$

Hence $I = \dfrac{{{e^{ax}}}}{{{a^2} - {b^2}}}\left( {\sin \left( {bx + c} \right) - b\cos \left( {bx + c} \right)} \right)$

Note:
It become a formula for these types of question form sine $I = \dfrac{{{e^{ax}}}}{{{a^2} - {b^2}}}\left( {\sin \left( {bx + c} \right) - b\cos \left( {bx + c} \right)} \right)$ and for cosine it is $I = \dfrac{{{e^{ax}}}}{{{a^2} - {b^2}}}\left( {\cos \left( {bx + c} \right) + b\sin \left( {bx + c} \right)} \right)$. Formula of integration by part is $\int {UVdx = U\int {Vdx - \int {\dfrac{{dU}}{{dx}}\left( {\int {Vdx} } \right)dx} } } $ where U and V is selected according to ILATE.