Answer: (a) \(\dfrac{dy}{dx}=\cos x-x\sin x\); (b) \(y=x-2\pi\); (c) \(\dfrac12-\dfrac{\pi\sqrt3}{12}\).

(a) Use the product rule on \(y=x\cos x\):

\(\frac{dy}{dx}=1\cdot\cos x+x(-\sin x).\)

Therefore

\(\frac{dy}{dx}=\cos x-x\sin x.\)

(b) When \(x=\pi\),

\(y=\pi\cos\pi=-\pi.\)

So the point is \((\pi,-\pi)\).

The gradient of the tangent at \(x=\pi\) is

\(\cos\pi-\pi\sin\pi=-1-0=-1.\)

Hence the gradient of the normal is \(1\).

Using \(y-y_1=m(x-x_1)\),

\(y+\pi=1(x-\pi).\)

So the equation of the normal is

\(y=x-2\pi.\)

(c) From part (a),

\(\frac{d}{dx}(x\cos x)=\cos x-x\sin x.\)

Rearrange this to express \(x\sin x\):

\(x\sin x=\cos x-\frac{d}{dx}(x\cos x).\)

Therefore

\(\int x\sin x\,dx=\int\cos x\,dx-x\cos x =\sin x-x\cos x.\)

Now evaluate between \(0\) and \(\dfrac{\pi}{6}\):

\(\int_0^{\frac{\pi}{6}}x\sin x\,dx =\left[\sin x-x\cos x\right]_0^{\frac{\pi}{6}}.\)

So

\(\int_0^{\frac{\pi}{6}}x\sin x\,dx =\frac12-\frac{\pi}{6}\cdot\frac{\sqrt3}{2}.\)

Hence

\(\int_0^{\frac{\pi}{6}}x\sin x\,dx =\frac12-\frac{\pi\sqrt3}{12}.\)