(a) Find: ∫ e^x · sin 2 x d x

\[\begin{aligned} I & =\int_{\text {II }} \underset{\text { I }}{e^{x}} \underset{\text { II }}{\sin 2 x} d x \\ I & =\sin 2 x e^{x}-\int \underset{\text { I }}{2 \cos 2 x} \underset{\text { II }}{e^{x}} d x \\ & =e^{x} \sin 2 x-2\left[\cos 2 x e^{x}-\int(-2 \sin 2 x) e^{x} d x\right] \end{aligned} \] \[I=e^{x} \sin 2 x-2 \cos 2 x e^{x}-4 I \] \(\displaystyle 5 I=e^{x} \sin 2 x-2 \cos 2 x e^{x}\) \[\therefore I=\frac{1}{5} e^{x}[(\sin 2 x-2 \cos 2 x)]+C \] Or \[\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+2\right)} d x \] Let \(\displaystyle x^{2}=t, 2 x d x=d t\) \[\therefore \int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+2\right)} d x=\int \frac{d t}{(t+1)(t+2)} \]