使用分部积分法（或其它方法），求 $\displaystyle\int \frac{1}{y}\ln y\,dy$。 已知 $y>1$ 且 $\displaystyle\frac{dy}{dx}=\frac{y}{\ln y}$，求用 $x$ 表示 $y$ 的表达式。 证明 $\displaystyle\frac{d}{dx}\big[x\sin(\ln x)-x\cos(\ln x)\big]=2\sin(\ln x)$。 求 $\displaystyle\int_{e^{1/2}}^{e^{2/2}}\big|\sin(\ln x)\big|\,dx$ 的精确值。

1. Define a suitable substitution for the integral.

Let $u = \ln y$.
Then $\frac{du}{dy} = \frac{1}{y} \implies du = \frac{1}{y} \,dy$.

2. Transform the integral entirely into terms of $u$.

$\int \frac{1}{y} \ln y \,dy = \int u \,du$

3. Integrate with respect to $u$, then substitute back $y$ and add the constant of integration.

$\int u \,du = \frac{1}{2}u^2 + C$
$= \frac{1}{2}(\ln y)^2 + C$

1. Evaluate the integral from part (a)(i) to use in this part.

To find $\displaystyle\int \frac{1}{y}\ln y\,dy$:
Let $u = \ln y$, then $du = \frac{1}{y}dy$.
$\displaystyle\int u\,du = \frac{1}{2}u^2 = \frac{1}{2}(\ln y)^2$.

2. Separate the variables of the differential equation and integrate.

$\displaystyle\frac{dy}{dx} = \frac{y}{\ln y}$
$\displaystyle\int \frac{\ln y}{y} \,dy = \int 1 \,dx$
Using the result from (a)(i):
$\displaystyle\frac{1}{2}(\ln y)^2 = x + C$, where $C$ is an arbitrary constant.

3. Make $y$ the subject, using the condition $y>1$ to determine the correct sign of the square root.

$(\ln y)^2 = 2x + 2C$
$\ln y = \pm\sqrt{2x + 2C}$
Since it is given that $y > 1$, we know $\ln y > 0$. Thus, we must reject the negative root.
$\ln y = \sqrt{2x + 2C}$
$y = e^{\sqrt{2x + 2C}}$
Letting $c = 2C$, the general solution is $y = e^{\sqrt{2x + c}}$.

1. Apply the product rule and chain rule to differentiate both terms

$\begin{aligned} \frac{d}{dx}\big[x\sin(\ln x)-x\cos(\ln x)\big] &= \left[ 1 \cdot \sin(\ln x) + x \cdot \cos(\ln x) \cdot \frac{1}{x} \right] - \left[ 1 \cdot \cos(\ln x) + x \cdot (-\sin(\ln x)) \cdot \frac{1}{x} \right] \end{aligned}$

2. Simplify the resulting expression to show the required result

$\begin{aligned} \frac{d}{dx}\big[x\sin(\ln x)-x\cos(\ln x)\big] &= \big(\sin(\ln x) + \cos(\ln x)\big) - \big(\cos(\ln x) - \sin(\ln x)\big) \\ &= \sin(\ln x) + \cos(\ln x) - \cos(\ln x) + \sin(\ln x) \\ &= 2\sin(\ln x) \end{aligned}$

1. Determine the sign of $\sin(\ln x)$ on the interval $[e^{1/2}, e^{2/2}]$ to resolve the absolute value.

For $x \in [e^{1/2}, e^{2/2}]$ (which simplifies to $[e^{0.5}, e]$), the bounds for $\ln x$ are:
$\frac{1}{2} \le \ln x \le 1$
Since $0 < \frac{1}{2} \le \ln x \le 1 < \pi$, we know that $\sin(\ln x) > 0$ strictly within this interval.
Thus, the absolute value can be dropped:
$\big|\sin(\ln x)\big| = \sin(\ln x)$

2. Apply the given derivative from part (b)(i) to find the antiderivative of $\sin(\ln x)$.

From part (b)(i), we are given:
$\frac{d}{dx}\big[x\sin(\ln x)-x\cos(\ln x)\big]=2\sin(\ln x)$
Integrating both sides with respect to $x$ gives:
$\int \sin(\ln x)\,dx = \frac{1}{2}\big[x\sin(\ln x)-x\cos(\ln x)\big] + C$

3. Evaluate the definite integral using the limits.

$\int_{e^{1/2}}^{e^{2/2}}\big|\sin(\ln x)\big|\,dx = \int_{e^{1/2}}^{e} \sin(\ln x)\,dx$
$= \left[ \frac{1}{2}\big(x\sin(\ln x)-x\cos(\ln x)\big) \right]_{e^{1/2}}^{e}$
Substitute $x = e$ and $x = e^{1/2}$:
$= \frac{1}{2}\big(e\sin(1)-e\cos(1)\big) - \frac{1}{2}\big(e^{1/2}\sin(\tfrac{1}{2})-e^{1/2}\cos(\tfrac{1}{2})\big)$
$= \frac{e}{2}(\sin 1 - \cos 1) - \frac{\sqrt{e}}{2}(\sin 0.5 - \cos 0.5)$