2017 H2 Mathematics Paper 1 Question 7

Integration Techniques

Answers

(i)

sin⁡(2m−2n)x4m−4n−sin⁡(2m+2n)x4m+4n+C{\frac{\sin (2m-2n)x}{4m-4n}} \allowbreak {- \frac{\sin (2m+2n)x}{4m+4n} + C}

(ii)

π{\pi}

Full solutions

(i)

∫sin⁡2mxsin⁡2nx  dx=−12∫(cos⁡(2m+2n)x−cos⁡(2m−2n)x)  dx=−12(sin⁡(2m+2n)x2m+2n−sin⁡(2m−2n)x2m−2n)+C=sin⁡(2m−2n)x4m−4n−sin⁡(2m+2n)x4m+4n+C  ■\begin{align*} & \int \sin 2m x \sin 2n x \; \mathrm{d}x \\ & = -\frac{1}{2} \int \Big( \cos (2m+2n) x - \cos (2m-2n) x \Big) \; \mathrm{d}x \\ & = -\frac{1}{2} \Big( \frac{\sin (2m+2n)x}{2m+2n} - \frac{\sin (2m-2n)x}{2m-2n} \Big) + C \\ & = \frac{\sin (2m-2n)x}{4m-4n} - \frac{\sin (2m+2n)x}{4m+4n} + C \; \blacksquare \end{align*}

(ii)

∫0π(sin⁡2mx+sin⁡2nx)2  dx∫0π(sin⁡22mx+2sin⁡2mxsin⁡2nx+sin⁡22nx)  dx∫0π(1−cos⁡4mx2−cos⁡(2m+2n)x+cos⁡(2m−2n)x+1−cos⁡4nx2)  dx=[x2−sin⁡4mx8m−sin⁡(2m+2n)x2m+2n+sin⁡(2m−2n)x2m−2n=[++x2−sin⁡4nx8n]0π=π2−sin⁡4mπ8m−sin⁡(2m+2n)π2m+2n+sin⁡(2m−2n)π2m−2n+π2−sin⁡4nπ8n−0=π−0=π  ■\begin{align*} & \int_0^\pi (\sin 2mx + \sin 2nx)^2 \; \mathrm{d}x \\ & \int_0^\pi \Big( \sin^2 2mx + 2 \sin 2mx \sin 2nx + \sin^2 2nx \Big) \; \mathrm{d}x \\ & \int_0^\pi \Bigg( \frac{1-\cos 4mx}{2} - \cos (2m+2n)x + \cos(2m-2n)x \\ & \qquad \qquad + \frac{1-\cos 4nx}{2} \Bigg) \; \mathrm{d}x \\ & = \Bigg[ \frac{x}{2} - \frac{\sin 4mx}{8m} - \frac{\sin (2m+2n)x}{2m+2n} + \frac{\sin (2m-2n)x}{2m-2n} \\ & \phantom{= \Bigg[ +} + \frac{x}{2} - \frac{\sin 4nx}{8n} \Bigg]_0^\pi \\ & = \frac{\pi}{2} - \frac{\sin 4m\pi}{8m} - \frac{\sin (2m+2n)\pi}{2m+2n} + \frac{\sin(2m-2n)\pi}{2m-2n} \\ & \qquad \qquad + \frac{\pi}{2}- \frac{\sin 4n \pi}{8n} - 0 \\ & = \pi - 0 \\ & = \pi \; \blacksquare \end{align*}