Answer: (a) \(\frac{\mathrm dy}{\mathrm dx}=\sin2x+2x\cos2x\); (b) \(y=x\); (c) \(\frac{\pi\sqrt3-3}{12}\).

We start with the main method. Use the product, chain and standard trigonometric derivative rules carefully.

(a) The curve is

\(y=x\sin2x.\)

Use the product rule:

\(\frac{\mathrm dy}{\mathrm dx}=x(2\cos2x)+\sin2x.\)

So

\(\frac{\mathrm dy}{\mathrm dx}=\sin2x+2x\cos2x.\)

(b) At \(x=\frac\pi4\),

\(y=\frac\pi4\sin\frac\pi2=\frac\pi4.\)

The gradient at \(x=\frac\pi4\) is

\(\sin\frac\pi2+2\left(\frac\pi4\right)\cos\frac\pi2=1+0=1.\)

The tangent has gradient \(1\) and passes through \(\left(\frac\pi4,\frac\pi4\right)\), so

\(y-\frac\pi4=1\left(x-\frac\pi4\right).\)

Hence

\(y=x.\)

(c) From part (a),

\(\frac{\mathrm d}{\mathrm dx}(x\sin2x)=\sin2x+2x\cos2x.\)

Therefore

\(2x\cos2x=\frac{\mathrm d}{\mathrm dx}(x\sin2x)-\sin2x.\)

So

\(\int_0^{\pi/6}2x\cos2x\,\mathrm dx =\left[x\sin2x\right]_0^{\pi/6}-\int_0^{\pi/6}\sin2x\,\mathrm dx.\)

Since

\(\int\sin2x\,\mathrm dx=-\frac12\cos2x,\)

we get

\(\int_0^{\pi/6}2x\cos2x\,\mathrm dx =\left[x\sin2x+\frac12\cos2x\right]_0^{\pi/6}.\)

At \(x=\frac\pi6\), this is

\(\frac\pi6\sin\frac\pi3+\frac12\cos\frac\pi3 =\frac{\pi\sqrt3}{12}+\frac14.\)

At \(x=0\), this is

\(\frac12.\)

Therefore the exact value is

\(\frac{\pi\sqrt3}{12}+\frac14-\frac12 =\frac{\pi\sqrt3}{12}-\frac14 =\frac{\pi\sqrt3-3}{12}.\)

This completes the solution and gives the required result.