2023 H2 Mathematics Paper 1 Question 4

Integration Techniques

Answers

(a)

sin⁡(p+q)x2(p+q)+sin⁡(p−q)x2(p−q)+C.{{\frac{\sin (p+q)x}{2(p+q)} + \frac{\sin (p-q)x}{2(p-q)} + C.}}

(b)

xsin⁡nxn+cos⁡nxn2+c.{{\frac{x \sin nx}{n} + \frac{\cos nx}{n^2} + c.}}

(c)

k=−2{{k = -2}} or k=0.{{k = 0.}}

(d)

π4.{{\frac{\pi}{4}.}}

Full solutions

(a)

∫cos⁡pxcos⁡qx  dx=∫12(cos⁡(p+q)x+cos⁡(p−q)x)  dx=sin⁡(p+q)x2(p+q)+sin⁡(p−q)x2(p−q)+C  ■\begin{align*} & \int \cos px \cos qx \; \mathrm{d}x \\ & = \int \frac{1}{2} \left( \cos (p+q)x + \cos (p-q)x \right) \; \mathrm{d}x \\ & = \frac{\sin (p+q)x}{2(p+q)} + \frac{\sin (p-q)x}{2(p-q)} + C \; \blacksquare \end{align*}

(b)

∫xcos⁡nx  dx=xsin⁡nxn−∫(1)sin⁡nxn  dx=xsin⁡nxn+cos⁡nxn2+c  ■\begin{align*} & \int x \cos nx \; \mathrm{d}x \\ & = x \frac{\sin nx}{n} - \int (1) \frac{\sin nx}{n} \; \mathrm{d}x \\ & = \frac{x \sin nx}{n} + \frac{\cos nx}{n^2} + c \; \blacksquare \end{align*}

(c)

∫0πxcos⁡nx  dx=[xsin⁡nxn+cos⁡nxn2]0π=πsin⁡nπn+cos⁡nπn2−0−cos⁡0n2=cos⁡nπn2−1n2={−2n2if n is odd0if n is even\begin{align*} & \int_0^\pi x \cos nx \; \mathrm{d}x \\ & = \left[ \frac{x \sin nx}{n} + \frac{\cos nx}{n^2} \right]_0^\pi \\ & = \frac{\pi \sin n\pi}{n} + \frac{\cos n\pi}{n^2} - 0 - \frac{\cos 0}{n^2} \\ &= \frac{\cos n\pi}{n^2} - \frac{1}{n^2} \\ &= \begin{cases} - \frac{2}{n^2} & \text{if } n \text{ is odd} \\ 0 & \text{if } n \text{ is even} \end{cases} \end{align*}
k=−2  ■ork=0  ■{k = -2 \; \blacksquare \quad \text{or} \quad k = 0 \; \blacksquare }

(d)

∫0π2∣xcos⁡2x∣  dx=∫0π4∣xcos⁡2x∣  dx−∫π4π2∣xcos⁡2x∣  dx=[xsin⁡2x2+cos⁡2x4]0π4−[xsin⁡2x2+cos⁡2x4]π4π2=π4sin⁡π22+cos⁡π24−0−14−(π2sin⁡π2+cos⁡π4−π4sin⁡π22−cos⁡π24)=π8−14+14+π8=π4  ■\begin{align*} & \int_0^{\frac{\pi}{2}} \left| x \cos 2x \right| \; \mathrm{d}x \\ & = \int_0^{\frac{\pi}{4}} \left| x \cos 2x \right| \; \mathrm{d}x - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left| x \cos 2x \right| \; \mathrm{d}x \\ & = \left[ \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} \right]_0^{\frac{\pi}{4}} - \left[ \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \\ & = \frac{\frac{\pi}{4} \sin \frac{\pi}{2}}{2} + \frac{\cos \frac{\pi}{2}}{4} - 0 - \frac{1}{4} \\ & \qquad - \left( \frac{\frac{\pi}{2} \sin \pi}{2} + \frac{\cos \pi}{4} - \frac{\frac{\pi}{4} \sin \frac{\pi}{2}}{2} - \frac{\cos \frac{\pi}{2}}{4} \right) \\ & = \frac{\pi}{8} - \frac{1}{4} + \frac{1}{4} + \frac{\pi}{8} \\ & = \frac{\pi}{4}\; \blacksquare \end{align*}

Question Commentary

This question is a mishmash of integration techniques involving trigonometric functions.

Part (a) uses the reverse of the factor formula, last seen just last year in 2022 P2 Q1, while (b) involves integration by parts, last seen in 2020 P1 Q7.

The last step of part (c) involves odd and even considerations last seen in 2016 P2 Q2, while (d) is about using the splitting of limits to tackle integration involving the modulus function.