2016 H2 Mathematics Paper 2 Question 2

Integration Techniques

Answers

(ai)
1nx2sin⁡nx+2n2xcos⁡nx−2n3sin⁡nx+C{\frac{1}{n} x^{2} \sin nx}\allowbreak {+ \frac{2}{n^2}x \cos nx}\allowbreak {- \frac{2}{n^3} \sin nx + C}
(aii)
{2πn2 if n is even6πn2 if n is odd{\begin{cases} 2\frac{\pi}{n^2} \quad &\textrm{ if } n \textrm{ is even} \\ 6\frac{\pi}{n^2} \quad &\textrm{ if } n \textrm{ is odd} \\ \end{cases}}

(b)

12π(45+ln⁡59) units3{\frac{1}{2} \pi \left( \frac{4}{5} + \ln \frac{5}{9} \right) \textrm{ units}^3}

Full solutions

(ai)
∫x2cos⁡nx  dx=1nx2sin⁡nx−∫1n2xsin⁡nx  dx=1nx2sin⁡nx−(−1n22xcos⁡nx−∫−1n22cos⁡nx  dx)=1nx2sin⁡nx+2n2xcos⁡nx−∫2n2cos⁡nx  dx=1nx2sin⁡nx+2n2xcos⁡nx−2n3sin⁡nx+C  ■\begin{align*} & \int x^{2} \cos nx \; \mathrm{d}x \\ & = \frac{1}{n} x^{2} \sin nx - \int \frac{1}{n} 2 x \sin nx \; \mathrm{d}x \\ & = \frac{1}{n} x^{2} \sin nx - \left( -\frac{1}{n^2}2 x \cos nx - \int - \frac{1}{n^2} 2 \cos nx \; \mathrm{d}x \right) \\ & = \frac{1}{n} x^{2} \sin nx + \frac{2}{n^2}x \cos nx - \int \frac{2}{n^2} \cos nx \; \mathrm{d}x \\ & = \frac{1}{n} x^{2} \sin nx + \frac{2}{n^2}x \cos nx - \frac{2}{n^3} \sin nx + C \; \blacksquare \end{align*}
(aii)
∫π2πx2cos⁡nx  dx=[1nx2sin⁡nx+2n2xcos⁡nx−2n3sin⁡nx]π2π=4π2nsin⁡2nπ+4πn2cos⁡2nπ−2n3sin⁡2nπ=+−(π2nsin⁡nπ+2πn2cos⁡nπ−2n3sin⁡nπ)=0+4πn2(1)−0−0−2πn2cos⁡nπ−0=4πn2−2πn2cos⁡nπ={4πn2−2πn2(1) if n is even4πn2−2πn2(−1) if n is odd={2πn2 if n is even6πn2 if n is odd  ■\begin{align*} & \int_\pi^{2\pi} x^{2} \cos nx \; \mathrm{d}x \\ & = \left[ \frac{1}{n} x^{2} \sin nx + \frac{2}{n^2}x \cos nx - \frac{2}{n^3} \sin nx \right]_\pi^{2\pi} \\ & = \frac{4\pi^2}{n} \sin 2n\pi + \frac{4\pi}{n^2} \cos 2n \pi - \frac{2}{n^3 \sin 2n\pi} \\ & \phantom{ = +} - \left( \frac{\pi^2}{n} \sin n\pi + \frac{2\pi}{n^2} \cos n \pi - \frac{2}{n^3 \sin n\pi} \right) \\ & = 0 + \frac{4\pi}{n^2}(1) - 0 - 0 - \frac{2\pi}{n^2} \cos n\pi - 0 \\ & = \frac{4\pi}{n^2} - \frac{2\pi}{n^2} \cos n\pi \\ & = \begin{cases} \frac{4\pi}{n^2} - \frac{2\pi}{n^2} (1) \quad &\textrm{ if } n \textrm{ is even} \\ \frac{4\pi}{n^2} - \frac{2\pi}{n^2} (-1) \quad &\textrm{ if } n \textrm{ is odd} \\ \end{cases} \\ & = \begin{cases} 2\frac{\pi}{n^2} \quad &\textrm{ if } n \textrm{ is even} \\ 6\frac{\pi}{n^2} \quad &\textrm{ if } n \textrm{ is odd} \\ \end{cases} \; \blacksquare \end{align*}

(b)

dudx=−2x\frac{\mathrm{d}u}{\mathrm{d}x} = -2x
When x=0,  u=9.{x=0, \; u = 9. \quad}When x=2,  u=5{x=2, \; u = 5}
Volume of solid obtained=π∫02(xx9−x2)2  dx=π∫95x3u21−2x  du=−12π∫959−uu2  du=12π∫599u2−1u  du=12π[−9u−ln⁡∣u∣]59=12π(−1−ln⁡9+95+ln⁡5)=12π(45+ln⁡59) units3  ■\begin{align*} & \textrm{Volume of solid obtained} \\ & = \pi \int_0^2 \left( \frac{x\sqrt{x}}{9-x^2} \right)^2 \; \mathrm{d}x \\ & = \pi \int_9^5 \frac{x^3}{u^2} \frac{1}{-2x} \; \mathrm{d}u \\ & = -\frac{1}{2} \pi \int_9^5 \frac{9-u}{u^2} \; \mathrm{d}u \\ & = \frac{1}{2} \pi \int_5^9 \frac{9}{u^2} - \frac{1}{u} \; \mathrm{d}u \\ & = \frac{1}{2} \pi \left[ - \frac{9}{u} - \ln \left|u\right| \right]_5^9 \\ & = \frac{1}{2} \pi \left( -1 - \ln 9 + \frac{9}{5} + \ln 5 \right) \\ & = \frac{1}{2} \pi \left( \frac{4}{5} + \ln \frac{5}{9} \right) \textrm{ units}^3 \; \blacksquare \end{align*}