Write a value of $\int \mathrm{e}^{2 x} \sin b x d x$
we know $\int f(x) g(x)=f(x) \int g(x)-\int f^{\prime}(x) \int g(x)$
Let $\int e^{a x} \sin b x d x=i$
Given that $\int e^{a x} \sin b x d x$
$i=\sin b x \int e^{a x}-\int b \cos b x \int e^{a x}$
$i=\sin b x \frac{e^{a x}}{a}-\int b \cos b x \frac{e^{a x}}{a}$
$i=\sin b x \frac{e^{a x}}{a}-\frac{1}{a}\left[b \cos b x \frac{e^{a x}}{a}-\frac{b^{2}}{a} \int e^{a x} \sin b x d x\right]$
$\mathrm{i}=\sin b x \frac{e^{a x}}{a}-\frac{b}{a^{2}} \cos b x e^{a x}+\frac{b^{2}}{a^{2}} \mathrm{i}$
$\mathrm{i}\left(1-\frac{b^{2}}{a^{2}}\right)=\frac{a \sin b x e^{a x}-b \cos b x e^{a x}}{a^{2}}$
$\mathrm{i}=\frac{a \sin b x e^{a x}-b \cos b x e^{a x}}{a^{2}}\left(\frac{a^{2}}{a^{2}-b^{2}}\right)$
$\int e^{a x} \sin b x d x=\frac{e^{a x}(a \sin b x-b \cos b x)}{a^{2}-b^{2}}$