求 $\int \sin 3x \cos x \text{ d}x$。 求 $\int \frac{x}{x^2 + 4x + 13} \text{ d}x$。 利用代换 $x = 3\sin\theta$ 求 $\int \sqrt{9 - x^2} \text{ d}x$。

1. Apply the product-to-sum trigonometric factor formula.

$\int \sin 3x \cos x \text{ d}x = \int \frac{1}{2} (\sin(3x + x) + \sin(3x - x)) \text{ d}x$
$= \frac{1}{2} \int (\sin 4x + \sin 2x) \text{ d}x$

2. Integrate the resulting terms and add the arbitrary constant of integration.

$= \frac{1}{2} \left( -\frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x \right) + c$
$= -\frac{1}{8}\cos 4x - \frac{1}{4}\cos 2x + c$

1. Express the numerator in terms of the derivative of the denominator.

$\int \frac{x}{x^2 + 4x + 13} \text{ d}x = \int \frac{\frac{1}{2}(2x + 4) - 2}{x^2 + 4x + 13} \text{ d}x$

2. Split the integral and evaluate the part involving the natural logarithm.

3. Complete the square for the denominator of the remaining integral.

$x^2 + 4x + 13 = (x + 2)^2 - 2^2 + 13 = (x + 2)^2 + 9$
$= \frac{1}{2} \ln(x^2 + 4x + 13) - 2 \int \frac{1}{(x + 2)^2 + 3^2} \text{ d}x$

4. Evaluate the inverse tangent integral and add the constant of integration.

1. Differentiate the given substitution to find $\text{d}x$ and substitute into the integral.

$x = 3\sin\theta \implies \frac{\text{d}x}{\text{d}\theta} = 3\cos\theta \implies \text{d}x = 3\cos\theta \text{ d}\theta$
$\int \sqrt{9 - x^2} \text{ d}x = \int \sqrt{9 - (3\sin\theta)^2} (3\cos\theta) \text{ d}\theta$

2. Simplify the expression under the square root using the Pythagorean identity.

$= \int \sqrt{9 - 9\sin^2\theta} (3\cos\theta) \text{ d}\theta$
$= \int \sqrt{9(1 - \sin^2\theta)} (3\cos\theta) \text{ d}\theta$
$= \int (3\cos\theta)(3\cos\theta) \text{ d}\theta$
$= \int 9\cos^2\theta \text{ d}\theta$

3. Integrate with respect to $\theta$ using the double-angle formula for cosine.

$= \int \frac{9}{2} (1 + \cos 2\theta) \text{ d}\theta$
$= \frac{9}{2} \left( \theta + \frac{1}{2}\sin 2\theta \right) + C$

4. Convert the result back to the original variable $x$ to find the final integral.

$= \frac{9}{2} \left( \theta + \sin\theta \cos\theta \right) + C$
Since $x = 3\sin\theta$, we have $\sin\theta = \frac{x}{3}$ and $\theta = \sin^{-1}\left(\frac{x}{3}\right)$.
Using the identity $\cos\theta = \sqrt{1 - \sin^2\theta}$ (since $\cos\theta \ge 0$ for principal values of $\sin^{-1}$):
$\cos\theta = \sqrt{1 - \left(\frac{x}{3}\right)^2} = \frac{\sqrt{9-x^2}}{3}$
Substituting these back into our expression:
$= \frac{9}{2} \left[ \sin^{-1}\left(\frac{x}{3}\right) + \left(\frac{x}{3}\right)\left(\frac{\sqrt{9-x^2}}{3}\right) \right] + C$
$= \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) + \frac{x}{2}\sqrt{9-x^2} + C$