It is given that $y = x\sin x + \cos^2\left(\tfrac{x}{2}\right).$

1. Differentiate the product $x\sin x$.

Using the product rule,
$\frac{d}{dx}(x\sin x)=x\cos x+\sin x.$

2. Differentiate $\cos^2\left(\frac{x}{2}\right)$ using the chain rule.

$=-\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right).$

3. Simplify the derivative into the required form $x\cos x+p\sin x$.

Since
$2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)=\sin x,$
we have
$-\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)=-\frac{1}{2}\sin x.$
Therefore,
$\frac{dy}{dx}=x\cos x+\sin x-\frac{1}{2}\sin x=x\cos x+\frac{1}{2}\sin x.$

1. Use the derivative from part (i) to express $2x\cos x$ in a useful form.

From part (i),
$\frac{dy}{dx}=x\cos x+\frac12\sin x.$
Hence,
$x\cos x=\frac{dy}{dx}-\frac12\sin x$
so
$2x\cos x=2\frac{dy}{dx}-\sin x.$

2. Integrate using the rewritten expression.

$\int_0^{\pi/2}2x\cos x\,dx=\int_0^{\pi/2}\left(2\frac{dy}{dx}-\sin x\right)dx$
$=\left[2y+\cos x\right]_0^{\pi/2}.$

3. Evaluate $y$ at the upper and lower limits using the given expression.

Given
$y=x\sin x+\cos^2\left(\frac{x}{2}\right).$
At $x=\frac{\pi}{2}$,
$y=\frac{\pi}{2}\sin\frac{\pi}{2}+\cos^2\frac{\pi}{4}$
$=\frac{\pi}{2}+\frac12.$
At $x=0$,
$y=0\sin0+\cos^2 0=1.$

4. Complete the definite integral and compare with $\pi+k$.

$\left[2y+\cos x\right]_0^{\pi/2}$
$=\left(2\left(\frac{\pi}{2}+\frac12\right)+\cos\frac{\pi}{2}\right)-\left(2(1)+\cos0\right)$
$=(\pi+1+0)-(2+1)$
$=\pi-2.$
Since
$\int_0^{\pi/2}2x\cos x\,dx=\pi+k,$
we have
$\pi+k=\pi-2,$
so
$k=-2.$