Answer: \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\cos2x-2x\sin2x\), and \(\displaystyle \int x\sin2x\,\mathrm{d}x=-\frac12x\cos2x+\frac14\sin2x+c\).

(a) Use the product rule on

\(y=x\cos2x\).

The derivative of \(x\) is \(1\), and the derivative of \(\cos2x\) is \(-2\sin2x\). Hence

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=1\cdot\cos2x+x(-2\sin2x)\).

So

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\cos2x-2x\sin2x\).

(b) From part (a),

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

Rearrange this to make \(x\sin2x\) the subject:

\(2x\sin2x=\cos2x-\dfrac{\mathrm{d}}{\mathrm{d}x}(x\cos2x)\),

so

\(x\sin2x=\frac12\cos2x-\frac12\dfrac{\mathrm{d}}{\mathrm{d}x}(x\cos2x)\).

Integrating both sides gives

\(\displaystyle \int x\sin2x\,\mathrm{d}x=\frac12\int\cos2x\,\mathrm{d}x-\frac12x\cos2x+c\).

Since \(\int\cos2x\,\mathrm{d}x=\frac12\sin2x\),

\(\displaystyle \int x\sin2x\,\mathrm{d}x=\frac14\sin2x-\frac12x\cos2x+c\).

Answer: (a) \(2\mathrm{e}^{3x-1}(\sin x+2\cos x)\). (b) \(\frac13\tan3x+C\).

(a) First rewrite the expression without a quotient:

\(\dfrac{\sin x+\cos x}{\mathrm e^{1-3x}}=\mathrm e^{3x-1}(\sin x+\cos x).\)

Use the product rule. Let

\(u=\mathrm e^{3x-1},\qquad v=\sin x+\cos x.\)

Then

\(u'=3\mathrm e^{3x-1},\qquad v'=\cos x-\sin x.\)

So

\(\dfrac{\mathrm d}{\mathrm dx}\left[\mathrm e^{3x-1}(\sin x+\cos x)\right] =3\mathrm e^{3x-1}(\sin x+\cos x)+\mathrm e^{3x-1}(\cos x-\sin x).\)

Factorising \(\mathrm e^{3x-1}\),

\(\mathrm e^{3x-1}\left(3\sin x+3\cos x+\cos x-\sin x\right) =\mathrm e^{3x-1}(2\sin x+4\cos x).\)

Therefore

\(\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{\sin x+\cos x}{\mathrm e^{1-3x}}\right)=2\mathrm e^{3x-1}(\sin x+2\cos x).\)

(b) Use the identity

\(1+\tan^2u=\sec^2u.\)

With \(u=3x\),

\(\int(1+\tan^2 3x)\,\mathrm dx=\int\sec^2 3x\,\mathrm dx.\)

Since \(\dfrac{\mathrm d}{\mathrm dx}(\tan3x)=3\sec^23x\), the integral is

\(\dfrac13\tan3x+C.\)