Using integration by parts, or otherwise, find $\displaystyle \int \frac{1}{y}\ln y\,dy$. Given that $y>1$ and $\displaystyle \frac{dy}{dx}=\frac{y}{\ln y}$, find $y$ in terms of $x$. Show that $\displaystyle \frac{d}{dx}\big[x\sin(\ln x)-x\cos(\ln x)\big]=2\sin(\ln x)$. Find the exact value of $\displaystyle \int_{e^{1/2}}^{e^{2}} |\sin(\ln x)|\,dx$.

1. Set up integration by parts

Let $I = \displaystyle \int \frac{1}{y}\ln y\,dy$.

Choose parts for integration:
Let $u = \ln y$ and $\frac{dv}{dy} = \frac{1}{y}$.
Then, differentiating $u$ and integrating $\frac{dv}{dy}$ gives:
$\frac{du}{dy} = \frac{1}{y}$ and $v = \ln y$.

2. Apply the integration by parts formula

Using the formula $\int u \, dv = uv - \int v \, du$:
$I = (\ln y)(\ln y) - \int (\ln y)\left(\frac{1}{y}\right)\,dy$
$I = (\ln y)^2 - I$

3. Rearrange to solve for the original integral

Add $I$ to both sides:
$2I = (\ln y)^2 + C'$
$I = \frac{1}{2}(\ln y)^2 + C$

(Note: This can also be solved elegantly "otherwise" by substituting $u = \ln y$, which gives $du = \frac{1}{y}dy$, directly yielding $\int u \, du = \frac{1}{2}u^2 + C = \frac{1}{2}(\ln y)^2 + C$.)

1. Separate the variables to set up the integrals.

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

2. Integrate both sides using the result from part (a)(i).

$\frac{1}{2}(\ln y)^2 = x + C$

3. Make $y$ the subject of the formula, applying the condition $y > 1$ to select the correct square root.

$(\ln y)^2 = 2x + 2C$
$\ln y = \pm\sqrt{2x + c} \quad \text{where } c = 2C$
Since $y > 1$, $\ln y > 0$, so we reject the negative root.
$\ln y = \sqrt{2x + c}$
$y = e^{\sqrt{2x + c}}$

1. Apply the product rule and chain rule to differentiate both terms with respect to $x$.

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

2. Simplify the resulting expression algebraically to obtain the required result.

$= [\sin(\ln x) + \cos(\ln x)] - [\cos(\ln x) - \sin(\ln x)]$
$= \sin(\ln x) + \cos(\ln x) - \cos(\ln x) + \sin(\ln x)$
$= 2\sin(\ln x)$

1. Justify the removal of the absolute value sign by checking the sign of $\sin(\ln x)$ on the given interval.

For $x \in [e^{1/2}, e^2]$, the values of $\ln x$ range continuously from $\frac{1}{2}$ to $2$.
Since $0 < \frac{1}{2} \le \ln x \le 2 < \pi$, it follows that $\sin(\ln x) > 0$ for the entire interval.
Thus, the absolute value can simply be dropped: $|\sin(\ln x)| = \sin(\ln x)$.

2. Use the result from part (b)(i) to set up the definite integral.

From part (b)(i), we know that $\frac{d}{dx}\big[x\sin(\ln x)-x\cos(\ln x)\big]=2\sin(\ln x)$, which implies:
$\int \sin(\ln x)\,dx = \frac{1}{2}\big[x\sin(\ln x)-x\cos(\ln x)\big]$
So, the definite integral becomes:
$\int_{e^{1/2}}^{e^{2}} \sin(\ln x)\,dx = \frac{1}{2}\big[x\sin(\ln x)-x\cos(\ln x)\big]_{e^{1/2}}^{e^2}$

3. Substitute the upper and lower limits to find the exact value.

$\begin{aligned} &\frac{1}{2}\big[x\sin(\ln x)-x\cos(\ln x)\big]_{e^{1/2}}^{e^2} \\ &= \frac{1}{2}\Big(e^2\sin(\ln e^2) - e^2\cos(\ln e^2)\Big) - \frac{1}{2}\Big(e^{1/2}\sin(\ln e^{1/2}) - e^{1/2}\cos(\ln e^{1/2})\Big) \\ &= \frac{1}{2}e^2(\sin 2 - \cos 2) - \frac{1}{2}e^{1/2}\left(\sin \frac{1}{2} - \cos \frac{1}{2}\right) \end{aligned}$