Find $\int \sin 3x\cos x\, dx$. Find $\displaystyle \int \frac{x}{x^2 + 4x + 13}\, dx$. Use the substitution $x = 3\sin \theta$ to find $\displaystyle \int \sqrt{9 - x^2}\, dx$.

1. Express the product of trigonometric functions as a sum using the factor formula.

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

2. Integrate the resulting expression term by term.

$\int \sin 3x \cos x \, dx = \frac{1}{2} \int (\sin 4x + \sin 2x) \, dx$\n$= \frac{1}{2} \left( -\frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x \right) + C$\n$= -\frac{1}{8}\cos 4x - \frac{1}{4}\cos 2x + C$

1. Express the numerator in terms of the derivative of the denominator to split the integral.

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

2. Integrate the first term to obtain a logarithmic function.

$\frac{1}{2} \int \frac{2x + 4}{x^2 + 4x + 13}\, dx = \frac{1}{2} \ln|x^2 + 4x + 13|$
Since $x^2 + 4x + 13 = (x+2)^2 + 9 > 0$ for all real $x$, the absolute value can be omitted:
$= \frac{1}{2} \ln(x^2 + 4x + 13)$

3. Complete the square for the denominator of the second term in preparation for inverse tangent integration.

$x^2 + 4x + 13 = (x + 2)^2 + 9$
$\int \frac{2}{x^2 + 4x + 13}\, dx = 2 \int \frac{1}{(x + 2)^2 + 3^2}\, dx$

4. Integrate the second term using the standard formula and state the final answer with the integration constant.

$2 \int \frac{1}{(x + 2)^2 + 3^2}\, dx = 2 \left[ \frac{1}{3} \tan^{-1}\left(\frac{x+2}{3}\right) \right] = \frac{2}{3} \tan^{-1}\left(\frac{x+2}{3}\right)$
Combining both parts:
$\int \frac{x}{x^2 + 4x + 13}\, dx = \frac{1}{2}\ln(x^2 + 4x + 13) - \frac{2}{3}\tan^{-1}\left(\frac{x+2}{3}\right) + C$

1. Differentiate the substitution to express $dx$ in terms of $d\theta$.

$x = 3\sin \theta$
$\frac{dx}{d\theta} = 3\cos \theta \implies dx = 3\cos \theta \, d\theta$

2. Substitute $x$ and $dx$ into the integral and simplify the integrand.

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

3. Apply the double-angle identity for $\cos^2 \theta$ to perform the integration.

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

4. Express $\sin \theta$, $\cos \theta$, and $\theta$ in terms of $x$ to substitute back, adding the constant of integration.

From $x = 3\sin \theta$, we have $\sin \theta = \frac{x}{3}$ and $\theta = \sin^{-1}\left(\frac{x}{3}\right)$.
Using $\cos^2 \theta = 1 - \sin^2 \theta$, we find $\cos \theta = \sqrt{1 - \left(\frac{x}{3}\right)^2} = \frac{\sqrt{9 - x^2}}{3}$.

Substituting these into the integrated expression:
$= \frac{9}{2} \left( \sin^{-1}\left(\frac{x}{3}\right) + \left(\frac{x}{3}\right)\left(\frac{\sqrt{9 - x^2}}{3}\right) \right) + C$
$= \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) + \frac{1}{2}x\sqrt{9 - x^2} + C$