Answer: (a) \(\frac{dy}{dx}=\sin x+x\cos x\); (b) \(y=x\); (c) \(x\sin x+\cos x+C\); (d) \(0.26\).
Answer: (a) \(\frac{dy}{dx}=\sin x+x\cos x\); (b) \(y=x\); (c) \(x\sin x+\cos x+C\); (d) \(0.26\).
(a) Use the product rule on \(y=x\sin x\):
\(\frac{dy}{dx}=1\cdot\sin x+x\cdot\cos x.\)
So
\(\frac{dy}{dx}=\sin x+x\cos x.\)
(b) At \(x=\frac{\pi}{2}\),
\(y=\frac{\pi}{2}\sin\frac{\pi}{2}=\frac{\pi}{2}.\)
The gradient is
\(\sin\frac{\pi}{2}+\frac{\pi}{2}\cos\frac{\pi}{2}=1.\)
The tangent has gradient \(1\) and passes through \(\left(\frac{\pi}{2},\frac{\pi}{2}\right)\), so
\(y-\frac{\pi}{2}=1\left(x-\frac{\pi}{2}\right).\)
Hence
\(y=x.\)
(c) From part (a),
\(\frac{d}{dx}(x\sin x)=\sin x+x\cos x.\)
Rearrange this to express \(x\cos x\):
\(x\cos x=\frac{d}{dx}(x\sin x)-\sin x.\)
Therefore
\(\int x\cos x\,dx=x\sin x-\int\sin x\,dx.\)
So
\(\int x\cos x\,dx=x\sin x+ \cos x+C.\)
(d) Use the result from part (c):
\(\int_0^{\frac{\pi}{4}}x\cos x\,dx =\left[x\sin x+\cos x\right]_0^{\frac{\pi}{4}}.\)
Thus
\(=\frac{\pi}{4}\sin\frac{\pi}{4}+\cos\frac{\pi}{4}-\cos0.\)
Since \(\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt2}{2}\),
\(=\frac{\pi\sqrt2}{8}+\frac{\sqrt2}{2}-1.\)
This gives
\(0.26\)
to 2 significant figures.
