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

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

Explanation

Solution

let us consider I=∫eax⋅sin⁡(bx+c)dxI = \int {{e^{ax}} \cdot \sin \left( {bx + c} \right)dx} then we will use integration by part ∫UVdx=U∫Vdx−∫dUdx(∫Vdx)dx\int {UVdx = U\int {Vdx - \int {\dfrac{{dU}}{{dx}}\left( {\int {Vdx} } \right)dx} } } where U=sin⁡(bx+c)U = \sin \left( {bx + c} \right) and V=eaxV = {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=∫eax⋅sin⁡(bx+c)dxI = \int {{e^{ax}} \cdot \sin \left( {bx + c} \right)dx}
Now, using integration by part ∫UVdx=U∫Vdx−∫dUdx(∫Vdx)dx\int {UVdx = U\int {Vdx - \int {\dfrac{{dU}}{{dx}}\left( {\int {Vdx} } \right)dx} } } where U=sin⁡(bx+c)U = \sin \left( {bx + c} \right) and V=eaxV = {e^{ax}}
So, I=sin⁡(bx+c)∫eaxdx−∫dsin⁡(bx+c)dx(∫eaxdx)dxI = \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} }
∫eaxdx=eaxa\int {{e^{ax}}dx = \dfrac{{{e^{ax}}}}{a}} and dsin⁡(bx+c)dx=bcos⁡(bx+c)\dfrac{{d\sin \left( {bx + c} \right)}}{{dx}} = b\cos \left( {bx + c} \right)
I=sin⁡(bx+c)eax−∫bcos⁡(bx+c)eaxdxaI = \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 ∫bcos⁡(bx+c)eaxdx\int {b\cos \left( {bx + c} \right){e^{ax}}dx} where U=cos⁡(bx+c)U = \cos \left( {bx + c} \right) and V=eaxV = {e^{ax}}
I=1a(sin⁡(bx+c)eax−(bcos⁡(bx+c)∫eaxdx−b∫dcos⁡(bx+c)dx(∫eaxdx)dx))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)
∫eaxdx=eaxa\int {{e^{ax}}dx = \dfrac{{{e^{ax}}}}{a}} and dcos⁡(bx+c)dx=−bsin⁡(bx+c)\dfrac{{d\cos \left( {bx + c} \right)}}{{dx}} = - b\sin \left( {bx + c} \right)
I=1a(sin⁡(bx+c)eax−(bacos⁡(bx+c)eax−b∫−basin⁡(bx+a)eaxdx))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=∫eax⋅sin⁡(bx+c)dxI = \int {{e^{ax}} \cdot \sin \left( {bx + c} \right)dx}
So, I=1asin⁡(bx+c)eax−(ba2cos⁡(bx+c)eax+(b2a2I))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)
a2−b2a2I=eaxa2(sin⁡(bx+c)−bcos⁡(bx+c))\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=eaxa2−b2(sin⁡(bx+c)−bcos⁡(bx+c))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=eaxa2−b2(sin⁡(bx+c)−bcos⁡(bx+c))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=eaxa2−b2(cos⁡(bx+c)+bsin⁡(bx+c))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 ∫UVdx=U∫Vdx−∫dUdx(∫Vdx)dx\int {UVdx = U\int {Vdx - \int {\dfrac{{dU}}{{dx}}\left( {\int {Vdx} } \right)dx} } } where U and V is selected according to ILATE.