There is one elegant method to solve integrals using matrice transformations which I saw in an old book. I don't think anyone has mentioned this technique.

Eg:

To evaluate $\int e^{ax}cos(bx+c)dx$, we define a matrix transformation to perform differentiation on a given vector

$$ \frac{d(V)}{dx} = \begin{bmatrix} a & -b\\ b & a \end{bmatrix} V $$ Such that $$ \frac{d}{dx} \begin{bmatrix} e^{ax}cos(bx+c)\\e^{ax}sin(bx+c) \end{bmatrix} = \begin{bmatrix} a & -b\\ b & a \end{bmatrix} \begin{bmatrix} e^{ax}cos(bx+c)\\e^{ax}sin(bx+c) \end{bmatrix} = \begin{bmatrix} ae^{ax}cos(bx+c)-be^{ax}sin(bx+c)\\ be^{ax}cos(bx+c)+ae^{ax}sin(bx+c) \end{bmatrix} $$

Now integration will be the inverse of this matrix $$ \therefore \int V dx = \begin{bmatrix}a & -b\\b & a\end{bmatrix}^{-1} = \frac{1}{a²+b²}\begin{bmatrix} a & b\\-b & a \end{bmatrix} $$

Now putting our original vector back, we get:

$$ \int \begin{bmatrix} e^{ax}cos(bx+c) \\ e^{ax}sin(bx+c) \end{bmatrix} dx = \frac{1}{a²+b²}\begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} e^{ax}cos(bx+c) \\ e^{ax}sin(bx+c) \end{bmatrix} = \frac{1}{a²+b²}\begin{bmatrix} ae^{ax}cos(bx+c)+be^{ax}sin(bx+c)\\ -be^{ax}cos(bx+c)+ae^{ax}sin(bx+c) \end{bmatrix} \therefore \begin{bmatrix} \int e^{ax}cos(bx+c)\\ \int e^{ax}sin(bx+c) \end{bmatrix} = \begin{bmatrix} \frac{1}{a²+b²}(ae^{ax}cos(bx+c)+be^{ax}sin(bx+c))\\ \frac{1}{a²+b²}(-be^{ax}cos(bx+c)+ae^{ax}sin(bx+c)) \end{bmatrix} $$

Equating corresponding values, we get:

$\int e^{ax}cos(bx+c)=\frac{e^{ax}}{a²+b²}(acos(bx+c)+bsin(bx+c))+const. \\ \& \int e^{ax}sin(bx+c)=\frac{e^{ax}}{a²+b²}(-bcos(bx+c)+asin(bx+c))+const.$

However, I haven't seen any other examples where you can use this technique; would love to know if there are more.

There is one elegant method to solve integrals using matrice transformations which I saw in an old book. I don't think anyone has mentioned this technique.

Eg:

To evaluate $\int e^{ax}\cos(bx+c)\mathrm dx$, we define a matrix transformation to perform differentiation on a given vector

$$ \frac{\mathrm d(V)}{\mathrm dx} = \begin{bmatrix} a & -b\\ b & a \end{bmatrix} V $$ Such that $$ \frac{\mathrm d}{\mathrm dx} \begin{bmatrix} e^{ax}\cos(bx+c)\\e^{ax}\sin(bx+c) \end{bmatrix} = \begin{bmatrix} a & -b\\ b & a \end{bmatrix} \begin{bmatrix} e^{ax}\cos(bx+c)\\e^{ax}\sin(bx+c) \end{bmatrix} = \begin{bmatrix} ae^{ax}\cos(bx+c)-be^{ax}\sin(bx+c)\\ be^{ax}\cos(bx+c)+ae^{ax}\sin(bx+c) \end{bmatrix} $$

Now, integration will be the inverse of this matrix $$ \therefore \int V\mathrm dx = \begin{bmatrix}a & -b\\b & a\end{bmatrix}^{-1} = \frac1{a^2+b^2}\begin{bmatrix} a & b\\-b & a \end{bmatrix} $$

Now, putting our original vector back, we get:

$$ \int \begin{bmatrix} e^{ax}\cos(bx+c) \\ e^{ax}\sin(bx+c) \end{bmatrix}\mathrm dx = \frac1{a^2+b^2}\begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} e^{ax}\cos(bx+c) \\ e^{ax}\sin(bx+c) \end{bmatrix} = \frac1{a^2+b^2}\begin{bmatrix} ae^{ax}\cos(bx+c)+be^{ax}\sin(bx+c)\\ -be^{ax}\cos(bx+c)+ae^{ax}\sin(bx+c) \end{bmatrix}$$ $$\therefore \begin{bmatrix} \displaystyle\int e^{ax}\cos(bx+c)\mathrm dx\\\displaystyle\int e^{ax}\sin(bx+c)\mathrm dx \end{bmatrix} = \begin{bmatrix} \frac1{a^2+b^2}(ae^{ax}\cos(bx+c)+be^{ax}\sin(bx+c))\\ \frac1{a^2+b^2}(-be^{ax}\cos(bx+c)+ae^{ax}\sin(bx+c)) \end{bmatrix} $$

Equating the corresponding elements, we get:

$$\int e^{ax}\cos(bx+c)\mathrm dx=\frac{e^{ax}}{a^2+b^2}(a\cos(bx+c)+b\sin(bx+c))+const. \\ \int e^{ax}\sin(bx+c)=\frac{e^{ax}}{a^2+b^2}(-b\cos(bx+c)+a\sin(bx+c))+const.$$

However, I haven't seen any other examples where you can use this technique; would love to know if there are more.