求 $\int \frac{3 - 2x}{\sqrt{5 + 4x - x^2}} \, \mathrm{d}x$。 证明
$\int_{\sqrt{2}}^{4} |x - 3| \, \mathrm{d}x = a - b\sqrt{2},$
其中 $a$ 和 $b$ 为待确定常数。 利用换元 $x = \sin u$，求 $\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}} \frac{1 + x^2}{\sqrt{1 - x^2}} \, \mathrm{d}x$ 的精确值。

1. Express the numerator in terms of the derivative of the quadratic expression inside the square root.

$\frac{\mathrm{d}}{\mathrm{d}x} (5 + 4x - x^2) = 4 - 2x$
Rewrite the integral by splitting the numerator:
$\int \frac{3 - 2x}{\sqrt{5 + 4x - x^2}} \, \mathrm{d}x = \int \frac{(4 - 2x) - 1}{\sqrt{5 + 4x - x^2}} \, \mathrm{d}x$
$= \int \frac{4 - 2x}{\sqrt{5 + 4x - x^2}} \, \mathrm{d}x - \int \frac{1}{\sqrt{5 + 4x - x^2}} \, \mathrm{d}x$

2. Integrate the first term using the standard result for $f'(x)[f(x)]^n$.

$\int (4 - 2x)(5 + 4x - x^2)^{-\frac{1}{2}} \, \mathrm{d}x = \frac{(5 + 4x - x^2)^{\frac{1}{2}}}{\frac{1}{2}}$
$= 2\sqrt{5 + 4x - x^2}$

3. Complete the square for the quadratic expression to prepare the second term for integration.

$5 + 4x - x^2 = 5 - (x^2 - 4x)$
$= 5 - \left[ (x - 2)^2 - 4 \right]$
$= 9 - (x - 2)^2$
$= 3^2 - (x - 2)^2$

4. Evaluate the second integral using the standard inverse sine formula and combine the results.

$\int \frac{1}{\sqrt{3^2 - (x - 2)^2}} \, \mathrm{d}x = \sin^{-1}\left(\frac{x - 2}{3}\right)$
Combine the two integrated parts and add the constant of integration:
$\int \frac{3 - 2x}{\sqrt{5 + 4x - x^2}} \, \mathrm{d}x = 2\sqrt{5 + 4x - x^2} - \sin^{-1}\left(\frac{x - 2}{3}\right) + C$

1. Split the integral at the critical point where the argument of the modulus is zero ($x = 3$).

$\int_{\sqrt{2}}^{4} |x - 3| \, \mathrm{d}x = \int_{\sqrt{2}}^{3} -(x - 3) \, \mathrm{d}x + \int_{3}^{4} (x - 3) \, \mathrm{d}x$
$= \int_{\sqrt{2}}^{3} (3 - x) \, \mathrm{d}x + \int_{3}^{4} (x - 3) \, \mathrm{d}x$

2. Integrate the linear expressions.

$= \left[ 3x - \frac{x^2}{2} \right]_{\sqrt{2}}^{3} + \left[ \frac{x^2}{2} - 3x \right]_{3}^{4}$

3. Substitute the limits and simplify to find the values of $a$ and $b$.

$= \left( 9 - \frac{9}{2} \right) - \left( 3\sqrt{2} - \frac{(\sqrt{2})^2}{2} \right) + \left( \frac{4^2}{2} - 12 \right) - \left( \frac{3^2}{2} - 9 \right)$
$= \frac{9}{2} - \left( 3\sqrt{2} - \frac{2}{2} \right) + \left( 8 - 12 \right) - \left( \frac{9}{2} - 9 \right)$
$= \frac{9}{2} - 3\sqrt{2} + 1 - 4 + \frac{9}{2}$
$= 6 - 3\sqrt{2}$

Comparing with $a - b\sqrt{2}$, we determine that $a = 6$ and $b = 3$.

1. Find the differential $\mathrm{d}x$ and the new limits of integration.

$x = \sin u \implies \frac{\mathrm{d}x}{\mathrm{d}u} = \cos u \implies \mathrm{d}x = \cos u \, \mathrm{d}u$

When $x = \frac{\sqrt{2}}{2}$, $\sin u = \frac{\sqrt{2}}{2} \implies u = \frac{\pi}{4}$
When $x = \frac{\sqrt{3}}{2}$, $\sin u = \frac{\sqrt{3}}{2} \implies u = \frac{\pi}{3}$

2. Substitute $x$, $\mathrm{d}x$ and the new limits into the integral to express it entirely in terms of $u$.

$\begin{aligned} \int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}} \frac{1 + x^2}{\sqrt{1 - x^2}} \, \mathrm{d}x &= \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1 + \sin^2 u}{\sqrt{1 - \sin^2 u}} (\cos u) \, \mathrm{d}u \\ &= \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1 + \sin^2 u}{\cos u} (\cos u) \, \mathrm{d}u \\ &= \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (1 + \sin^2 u) \, \mathrm{d}u \end{aligned}$

3. Use the double angle identity for $\sin^2 u$ to integrate the expression.

$\begin{aligned} \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (1 + \sin^2 u) \, \mathrm{d}u &= \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \left(1 + \frac{1 - \cos 2u}{2}\right) \, \mathrm{d}u \\ &= \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \left(\frac{3}{2} - \frac{1}{2}\cos 2u\right) \, \mathrm{d}u \\ &= \left[ \frac{3}{2}u - \frac{1}{4}\sin 2u \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} \end{aligned}$

4. Evaluate the definite integral using the upper and lower limits.

$\begin{aligned} \left[ \frac{3}{2}u - \frac{1}{4}\sin 2u \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} &= \left( \frac{3}{2}\left(\frac{\pi}{3}\right) - \frac{1}{4}\sin\left(\frac{2\pi}{3}\right) \right) - \left( \frac{3}{2}\left(\frac{\pi}{4}\right) - \frac{1}{4}\sin\left(\frac{2\pi}{4}\right) \right) \\ &= \left( \frac{\pi}{2} - \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) \right) - \left( \frac{3\pi}{8} - \frac{1}{4}(1) \right) \\ &= \frac{\pi}{2} - \frac{\sqrt{3}}{8} - \frac{3\pi}{8} + \frac{1}{4} \\ &= \frac{4\pi}{8} - \frac{3\pi}{8} + \frac{2}{8} - \frac{\sqrt{3}}{8} \\ &= \frac{\pi + 2 - \sqrt{3}}{8} \end{aligned}$