求 $\displaystyle \int \sin 3x \cos x\, dx$。 求 $\displaystyle \int e^{2x} \cos 3x\, dx$。

1. Express the product of sine and cosine as a sum using the factor formula.

$\sin 3x \cos x = \frac{1}{2}[\sin(3x + x) + \sin(3x - x)]$
$= \frac{1}{2}(\sin 4x + \sin 2x)$

2. Integrate the resulting terms using standard integration rules.

$\int \frac{1}{2}(\sin 4x + \sin 2x) \, dx = \frac{1}{2} \left( -\frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x \right) + C$
$= -\frac{1}{8}\cos 4x - \frac{1}{4}\cos 2x + C$

1. Apply integration by parts to the original integral by choosing $u = e^{2x}$ and $\frac{dv}{dx} = \cos 3x$.

Let $I = \displaystyle \int e^{2x} \cos 3x \, dx$.

Using the integration by parts formula $\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx$:
Let $u = e^{2x} \implies \frac{du}{dx} = 2e^{2x}$
Let $\frac{dv}{dx} = \cos 3x \implies v = \frac{1}{3}\sin 3x$

$I = \frac{1}{3}e^{2x}\sin 3x - \int \frac{2}{3}e^{2x}\sin 3x \, dx$

2. Apply integration by parts a second time to the new integral $\int e^{2x}\sin 3x \, dx$.

For $\int e^{2x}\sin 3x \, dx$:
Let $u = e^{2x} \implies \frac{du}{dx} = 2e^{2x}$
Let $\frac{dv}{dx} = \sin 3x \implies v = -\frac{1}{3}\cos 3x$

$\int e^{2x}\sin 3x \, dx = -\frac{1}{3}e^{2x}\cos 3x - \int \left(-\frac{1}{3}\cos 3x\right)(2e^{2x}) \, dx$
$= -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{3} \int e^{2x}\cos 3x \, dx$
$= -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{3} I$

3. Substitute the result back into the original expression to form an algebraic equation in terms of $I$.

Substitute this back into the initial expansion for $I$:
$I = \frac{1}{3}e^{2x}\sin 3x - \frac{2}{3} \left( -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{3} I \right)$
$I = \frac{1}{3}e^{2x}\sin 3x + \frac{2}{9}e^{2x}\cos 3x - \frac{4}{9}I$

4. Rearrange to solve for $I$ and include the arbitrary constant of integration $C$.

$I + \frac{4}{9}I = \frac{1}{3}e^{2x}\sin 3x + \frac{2}{9}e^{2x}\cos 3x$
$\frac{13}{9}I = \frac{1}{3}e^{2x}\sin 3x + \frac{2}{9}e^{2x}\cos 3x$
$I = \frac{9}{13} \left( \frac{1}{3}e^{2x}\sin 3x + \frac{2}{9}e^{2x}\cos 3x \right) + C$
$I = \frac{3}{13}e^{2x}\sin 3x + \frac{2}{13}e^{2x}\cos 3x + C$