Use identities for integration and geometry for the regions · source reference

Printed question

(a) Find $\int(\cot^62x+\cot^42x-\sin(x/2)\sin(3x/2))dx$. [3]

(b) The diagram shows $y=e^{2x}$ and $y=(x+1)^2$, intersection $(-1.2785,0.07754)$ and $y$-intercept 1. R is the finite region bounded by both curves. Find the volume when R is rotated $2\pi$ about the $y$-axis, to 4 decimal places. [3]

(c) C: $x=2(\sin(t/2)-\cos(t/2))$, $y=\sin(t/2)+\cos(t/2)$, $-3\pi/2\le t\le-\pi/2$. Line L: $y=-2x-5$, meets C at $(-2,-1)$. S is enclosed by L,C and the $x$-axis as shown. (i) Show C’s $x$-intercept is $-2\sqrt2$. [2] (ii) Find the exact area of S. [6]

[2020/ASRJC/Prelim/II/4]

Printed working

(a)

$$\begin{aligned}\int(\cot^62x+\cot^42x-\sin(x/2)\sin(3x/2))dx&=\int[\cot^42x(\cot^22x+1)+(\cos2x-\cos x)/2]dx\\&=\int[\cot^42x\csc^22x+(\cos2x-\cos x)/2]dx\\&=-\frac{\cot^52x}{10}+\frac{\sin2x}4-\frac{\sin x}2+c.\end{aligned}$$

(b)

The working calls the same region “S”.

$$V=\pi\int_{0.077542}^{1}(\tfrac12\ln y)^2dy+\pi\int_0^{0.077542}(-1-\sqrt y)^2dy-\pi\int_0^1(-1+\sqrt y)^2dy=0.5593\quad\text{units}^3\ (4\text{ d.p.}).$$

(c)(i)

When $y=0$, $0=\sin(t/2)+\cos(t/2)$, so $\tan(t/2)=-1$ and $t=-\pi/2$ in the specified domain.

$$x=2[\sin(-\pi/4)-\cos(-\pi/4)]=2(-1/\sqrt2-1/\sqrt2)=-2\sqrt2.$$

(c)(ii)

$$\begin{aligned}\text{Area}&=-\int_{-2\sqrt2}^{-2}y_Cdx-\frac12(1/2)(1)\\&=-\int_{-\pi/2}^{-\pi}[\sin(t/2)+\cos(t/2)][2(1/2)(\sin(t/2)+\cos(t/2))]dt-\frac14\\&=\int_{-\pi}^{-\pi/2}(\sin(t/2)+\cos(t/2))^2dt-\frac14\\&=\int_{-\pi}^{-\pi/2}[\sin^2(t/2)+\cos^2(t/2)+2\sin(t/2)\cos(t/2)]dt-\frac14\\&=\int_{-\pi}^{-\pi/2}(1+\sin t)dt-\frac14=[t-\cos t]_{-\pi}^{-\pi/2}-\frac14\\&=-\pi/2-(-\pi-\cos(-\pi))-\frac14=\frac\pi2-\frac54\quad\text{units}^2.\end{aligned}$$

Teacher notes

From MF26: $\cos P-\cos Q=-2\sin((P+Q)/2)\sin((P-Q)/2)$. Thus $-2\sin(mx)\sin(nx)=\cos((m+n)x)-\cos((m-n)x)$, replacing $m$ by $3/2$, $n$ by $1/2$.

(c)(ii) $\int_{-2\sqrt2}^{-2}y_Cdx<0$ because C lies below the $x$-axis on $[-2\sqrt2,-2]$. Native diagrams retain the branches $x=-1\pm\sqrt y$, $y=e^{2x}$, level 1, rotation arrow, and the area diagram’s $-2\sqrt2,-5/2,-2,C,S,L,(-2,-1)$ labels.

Printed answer key

(a) $-\cot^52x/10+\sin2x/4-\sin x/2+c$; (b) $0.5593$ units$^3$ (4 d.p.); (c)(ii) $\pi/2-5/4$ units$^2$. No (c)(i) short answer is printed.

Source diagrams in native form