求常数 $p$ 的精确值，使得
$\displaystyle \int_{5}^{p+9} \frac{1}{\sqrt{9-x}}\,dx = \displaystyle \int_{0}^{\tfrac{1}{4}} \frac{1}{\sqrt{1-4x^{2}}}\,dx.$ 利用换元 $x = \cos^{2}\theta$ 求
$\displaystyle \int_{0}^{\tfrac{1}{2}} \frac{\sqrt{x}}{\sqrt{1-x}}\,dx.$
的精确值。

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完整解法

8(a)

1. Evaluate the left-hand side integral.

$\int_{5}^{p+9} (9-x)^{-\frac{1}{2}} \,dx = \left[ \frac{(9-x)^{\frac{1}{2}}}{-\frac{1}{2}} \right]_{5}^{p+9}$
$= \left[ -2\sqrt{9-x} \right]_{5}^{p+9}$
$= -2\sqrt{9 - (p+9)} - \left( -2\sqrt{9-5} \right)$
$= -2\sqrt{-p} + 2\sqrt{4}$
$= 4 - 2\sqrt{-p}$

2. Evaluate the right-hand side integral.

$\int_{0}^{\frac{1}{4}} \frac{1}{\sqrt{1-4x^2}} \,dx = \int_{0}^{\frac{1}{4}} \frac{1}{\sqrt{1-(2x)^2}} \,dx$
$= \left[ \frac{1}{2} \sin^{-1}(2x) \right]_{0}^{\frac{1}{4}}$
$= \frac{1}{2} \left( \sin^{-1}\left(\frac{1}{2}\right) - \sin^{-1}(0) \right)$
$= \frac{1}{2} \left( \frac{\pi}{6} - 0 \right)$
$= \frac{\pi}{12}$

3. Equate the two evaluated expressions and solve for $p$.

$4 - 2\sqrt{-p} = \frac{\pi}{12}$
$2\sqrt{-p} = 4 - \frac{\pi}{12}$
$\sqrt{-p} = 2 - \frac{\pi}{24}$
$-p = \left( 2 - \frac{\pi}{24} \right)^2$
$p = -\left( 2 - \frac{\pi}{24} \right)^2$

答案

$p = -\left(2 - \frac{\pi}{24}\right)^2$

8(b)

1. Find the differential $dx$ and the new limits for $\theta$.

Using $x = \cos^2\theta$ :
$\frac{dx}{d\theta} = 2\cos\theta(-\sin\theta) = -2\cos\theta\sin\theta$

Finding the new limits:
When $x = 0$ , $\cos^2\theta = 0 \implies \cos\theta = 0 \implies \theta = \frac{\pi}{2}$
When $x = \frac{1}{2}$ , $\cos^2\theta = \frac{1}{2} \implies \cos\theta = \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4}$

2. Substitute $x$, $dx$, and the limits into the integral and simplify.

\begin{aligned} \int_{0}^{\frac{1}{2}} \frac{\sqrt{x}}{\sqrt{1-x}} \, dx &= \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \frac{\sqrt{\cos^2\theta}}{\sqrt{1-\cos^2\theta}} (-2\cos\theta\sin\theta) \, d\theta \\ &= \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \frac{\cos\theta}{\sqrt{\sin^2\theta}} (-2\cos\theta\sin\theta) \, d\theta \\ &= \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \frac{\cos\theta}{\sin\theta} (-2\cos\theta\sin\theta) \, d\theta \\ &= \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} -2\cos^2\theta \, d\theta \\ &= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2\cos^2\theta \, d\theta \end{aligned}

3. Use the double angle identity to integrate.

Applying the identity $2\cos^2\theta = 1 + \cos 2\theta$ :

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

4. Substitute the limits to find the exact value.

\begin{aligned} \left[ \theta + \frac{1}{2}\sin 2\theta \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} &= \left( \frac{\pi}{2} + \frac{1}{2}\sin(\pi) \right) - \left( \frac{\pi}{4} + \frac{1}{2}\sin\left(\frac{\pi}{2}\right) \right) \\ &= \left( \frac{\pi}{2} + 0 \right) - \left( \frac{\pi}{4} + \frac{1}{2}(1) \right) \\ &= \frac{\pi}{4} - \frac{1}{2} \end{aligned}

答案

$\dfrac{\pi}{4} - \dfrac{1}{2}$

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完整解法

8(a)

8(b)

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