通过将 $\sin^3 x$ 写成 $\sin^2 x(\sin x)$，求 $\int \sin^3 x \cos^3 x \text{ d}x$。 求 $\int \frac{5 - 6x}{\sqrt{5 - 3x^2 + 2x}} \text{ d}x$。

1. Use the given hint and the identity $\sin^2 x = 1 - \cos^2 x$ to rewrite the integrand.

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

2. Apply the substitution $u = \cos x$ to evaluate the integral.

Let $u = \cos x$. Then $\frac{\text{d}u}{\text{d}x} = -\sin x$, which gives $\sin x \text{ d}x = -\text{d}u$.
$\int (1 - \cos^2 x) \cos^3 x \sin x \text{ d}x = \int (1 - u^2) u^3 (-\text{d}u)$
$= \int (u^5 - u^3) \text{ d}u$
$= \frac{u^6}{6} - \frac{u^4}{4} + C$

3. Substitute back $u = \cos x$ to express the final answer in terms of $x$.

$= \frac{1}{6}\cos^6 x - \frac{1}{4}\cos^4 x + C$

1. Split the numerator to express it in terms of the derivative of the quadratic expression in the denominator.

Let $f(x) = 5 - 3x^2 + 2x$. Then $f'(x) = 2 - 6x$.
We rewrite the numerator $5 - 6x$ as $(2 - 6x) + 3$ to split the integral:
$\int \frac{5 - 6x}{\sqrt{5 - 3x^2 + 2x}} \text{ d}x = \int \frac{2 - 6x + 3}{\sqrt{5 - 3x^2 + 2x}} \text{ d}x$
$= \int \frac{2 - 6x}{\sqrt{5 - 3x^2 + 2x}} \text{ d}x + \int \frac{3}{\sqrt{5 - 3x^2 + 2x}} \text{ d}x$

2. Evaluate the first integral using the exact derivative.

Using the reverse chain rule for $\int f'(x)[f(x)]^n \text{ d}x$:
$\int (2 - 6x)(5 - 3x^2 + 2x)^{-\frac{1}{2}} \text{ d}x = \frac{(5 - 3x^2 + 2x)^{\frac{1}{2}}}{\frac{1}{2}} = 2\sqrt{5 - 3x^2 + 2x}$

3. Complete the square for the quadratic expression to prepare the second integral.

Factorising the coefficient of $x^2$ and completing the square:
$5 - 3x^2 + 2x = 5 - 3\left(x^2 - \frac{2}{3}x\right)$
$= 5 - 3\left[ \left(x - \frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 \right]$
$= 5 - 3\left(x - \frac{1}{3}\right)^2 + \frac{1}{3}$
$= \frac{16}{3} - 3\left(x - \frac{1}{3}\right)^2$
$= 3\left[ \frac{16}{9} - \left(x - \frac{1}{3}\right)^2 \right]$

4. Evaluate the second integral using the standard inverse sine formula and state the final answer.

$\int \frac{3}{\sqrt{5 - 3x^2 + 2x}} \text{ d}x = \int \frac{3}{\sqrt{3\left[ \frac{16}{9} - \left(x - \frac{1}{3}\right)^2 \right]}} \text{ d}x$
$= \frac{3}{\sqrt{3}} \int \frac{1}{\sqrt{\left(\frac{4}{3}\right)^2 - \left(x - \frac{1}{3}\right)^2}} \text{ d}x$
Using the standard formula $\int \frac{1}{\sqrt{a^2 - x^2}} \text{ d}x = \sin^{-1}\left(\frac{x}{a}\right)$:
$= \sqrt{3} \sin^{-1}\left( \frac{x - \frac{1}{3}}{\frac{4}{3}} \right)$
$= \sqrt{3} \sin^{-1}\left( \frac{3x - 1}{4} \right)$
Combining the results and adding the constant of integration $c$:
$\int \frac{5 - 6x}{\sqrt{5 - 3x^2 + 2x}} \text{ d}x = 2\sqrt{5 - 3x^2 + 2x} + \sqrt{3} \sin^{-1}\left( \frac{3x - 1}{4} \right) + c$