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IAL 2022 Jan Q3

A Level / Edexcel / P3

IAL 2022 Jan Paper · Question 3

题目

Problem

(i) Find, in simplest form,

∫(2x−5)7 dx\int(2x-5)^7\,dx
(2)

(ii) Show, by algebraic integration, that

∫0π/34sin⁡x1+2cos⁡x dx=ln⁡a\int_0^{\pi/3}\frac{4\sin x}{1+2\cos x}\,dx=\ln a

where aa is a rational constant to be found.

(4)
题目中文翻译

(i) 求

∫(2x−5)7 dx\int(2x-5)^7\,dx

的最简形式。

(ii) 用代数积分证明

∫0π/34sin⁡x1+2cos⁡x dx=ln⁡a\int_0^{\pi/3}\frac{4\sin x}{1+2\cos x}\,dx=\ln a

其中 aa 是需要求出的有理常数。

解答

(i)

We need to integrate

∫(2x−5)7 dx\int(2x-5)^7\,\mathrm{d}x

Using the reverse chain rule,

∫(2x−5)7 dx=(2x−5)88⋅12+c\int(2x-5)^7\,\mathrm{d}x =\frac{(2x-5)^8}{8}\cdot\frac12+c

So

∫(2x−5)7 dx=(2x−5)816+c\boxed{\int(2x-5)^7\,\mathrm{d}x =\frac{(2x-5)^8}{16}+c}

(ii)

Let

u=1+2cos⁡xu=1+2\cos x

Then

dudx=−2sin⁡x\frac{\mathrm{d}u}{\mathrm{d}x}=-2\sin x

So

4sin⁡x dx=−2 du4\sin x\,\mathrm{d}x=-2\,\mathrm{d}u

Therefore

∫4sin⁡x1+2cos⁡x dx=∫−2u du\int\frac{4\sin x}{1+2\cos x}\,\mathrm{d}x =\int -\frac{2}{u}\,\mathrm{d}u

Hence

∫4sin⁡x1+2cos⁡x dx=−2ln⁡u\int\frac{4\sin x}{1+2\cos x}\,\mathrm{d}x =-2\ln u

Substituting back,

∫4sin⁡x1+2cos⁡x dx=−2ln⁡(1+2cos⁡x)\int\frac{4\sin x}{1+2\cos x}\,\mathrm{d}x =-2\ln(1+2\cos x)

Now evaluate between 00 and π3\dfrac{\pi}{3}:

∫0π/34sin⁡x1+2cos⁡x dx=[−2ln⁡(1+2cos⁡x)]0π/3=−2ln⁡(1+2⋅12)−[−2ln⁡(1+2⋅1)]=−2ln⁡2+2ln⁡3\begin{aligned} \int_0^{\pi/3}\frac{4\sin x}{1+2\cos x}\,\mathrm{d}x &=\left[-2\ln(1+2\cos x)\right]_0^{\pi/3} \\ &=-2\ln\left(1+2\cdot\frac12\right) -\left[-2\ln(1+2\cdot1)\right] \\ &=-2\ln2+2\ln3 \end{aligned}

So

∫0π/34sin⁡x1+2cos⁡x dx=2ln⁡3−2ln⁡2\int_0^{\pi/3}\frac{4\sin x}{1+2\cos x}\,\mathrm{d}x =2\ln3-2\ln2

Therefore

2ln⁡3−2ln⁡2=ln⁡9−ln⁡4=ln⁡942\ln3-2\ln2 =\ln9-\ln4 =\ln\frac94

Hence

a=94\boxed{a=\frac94}