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IAL 2020 Oct Q5

A Level / Edexcel / P3

IAL 2020 Oct Paper · Question 5

题目

Problem

(a) Show that

sin⁡3x≡3sin⁡x−4sin⁡3x\sin 3x\equiv 3\sin x-4\sin^3x
(4)

(b) Hence find, using algebraic integration,

∫0π3sin⁡3x dx\int_0^{\frac{\pi}{3}} \sin^3x\,dx
(4)
题目中文翻译

(a) 证明

sin⁡3x≡3sin⁡x−4sin⁡3x\sin 3x\equiv 3\sin x-4\sin^3x

(b) 由此利用代数积分法求

∫0π3sin⁡3x dx\int_0^{\frac{\pi}{3}} \sin^3x\,dx

解答

(a)

Start with the left hand side:

sin⁡3x=sin⁡(2x+x)\sin3x=\sin(2x+x)

Use the compound angle formula:

sin⁡(2x+x)=sin⁡2xcos⁡x+cos⁡2xsin⁡x\sin(2x+x)=\sin2x\cos x+\cos2x\sin x

Now use

sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x

and

cos⁡2x=1−2sin⁡2x\cos2x=1-2\sin^2x

Then

sin⁡3x=2sin⁡xcos⁡xcos⁡x+(1−2sin⁡2x)sin⁡x=2sin⁡xcos⁡2x+sin⁡x−2sin⁡3x\begin{aligned} \sin3x &=2\sin x\cos x\cos x+(1-2\sin^2x)\sin x\\ &=2\sin x\cos^2x+\sin x-2\sin^3x \end{aligned}

Since

cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x

we get

sin⁡3x=2sin⁡x(1−sin⁡2x)+sin⁡x−2sin⁡3x=2sin⁡x−2sin⁡3x+sin⁡x−2sin⁡3x=3sin⁡x−4sin⁡3x\begin{aligned} \sin3x &=2\sin x(1-\sin^2x)+\sin x-2\sin^3x\\ &=2\sin x-2\sin^3x+\sin x-2\sin^3x\\ &=3\sin x-4\sin^3x \end{aligned}

Therefore

sin⁡3x≡3sin⁡x−4sin⁡3x\boxed{\sin3x\equiv 3\sin x-4\sin^3x}

(b)

From part (a),

sin⁡3x=3sin⁡x−4sin⁡3x\sin3x=3\sin x-4\sin^3x

Rearrange:

4sin⁡3x=3sin⁡x−sin⁡3x4\sin^3x=3\sin x-\sin3x

so

sin⁡3x=34sin⁡x−14sin⁡3x\sin^3x=\frac34\sin x-\frac14\sin3x

Therefore

∫0π3sin⁡3x dx=∫0π3(34sin⁡x−14sin⁡3x) dx\int_0^{\frac{\pi}{3}}\sin^3x\,\mathrm{d}x =\int_0^{\frac{\pi}{3}}\left(\frac34\sin x-\frac14\sin3x\right)\,\mathrm{d}x

Integrate:

∫(34sin⁡x−14sin⁡3x) dx=−34cos⁡x+112cos⁡3x\int \left(\frac34\sin x-\frac14\sin3x\right)\,\mathrm{d}x =-\frac34\cos x+\frac{1}{12}\cos3x

So

∫0π3sin⁡3x dx=[−34cos⁡x+112cos⁡3x]0π3=(−34cos⁡π3+112cos⁡π)−(−34cos⁡0+112cos⁡0)=(−38−112)−(−34+112)=−1124+1624=524\begin{aligned} \int_0^{\frac{\pi}{3}}\sin^3x\,\mathrm{d}x &=\left[-\frac34\cos x+\frac{1}{12}\cos3x\right]_0^{\frac{\pi}{3}}\\ &=\left(-\frac34\cos\frac{\pi}{3}+\frac{1}{12}\cos\pi\right) -\left(-\frac34\cos0+\frac{1}{12}\cos0\right)\\ &=\left(-\frac38-\frac{1}{12}\right)-\left(-\frac34+\frac{1}{12}\right)\\ &=-\frac{11}{24}+\frac{16}{24}\\ &=\frac{5}{24} \end{aligned}

Hence

524\boxed{\frac{5}{24}}