Choose the trigonometric branch before simplifying roots · source reference

Printed question

(a) Use $x=2\tan\theta$ to find the exact value of $\int_{2/\sqrt3}^{2}\frac1{x\sqrt{x^2+4}}dx$. [6]

(b)(i) By writing $x=A(1-x)+B$, where $A,B$ are constants, prove that

$$\int\frac x{\sqrt{1-x}}dx=\frac23(1-x)^{3/2}-2(1-x)^{1/2}+C,$$

where $C$ is an arbitrary constant. [2]

(ii) Hence find $\int\sqrt x(\sin^{-1}\sqrt x)dx$. [3]

[2009/SRJC/Prelim/I/9]

Printed working

(a)

$x=2\tan\theta$, $dx/d\theta=2\sec^2\theta$; $x=2$ gives $\theta=\pi/4$, $x=2/\sqrt3$ gives $\theta=\pi/6$.

$$\begin{aligned}\int_{2/\sqrt3}^2\frac{dx}{x\sqrt{x^2+4}}&=\int_{\pi/6}^{\pi/4}\frac{2\sec^2\theta}{(2\tan\theta)\sqrt{4\tan^2\theta+4}}d\theta\\&=\int_{\pi/6}^{\pi/4}\frac{2\sec^2\theta}{(2\tan\theta)\sqrt{4(\tan^2\theta+1)}}d\theta\\&=\int_{\pi/6}^{\pi/4}\frac{2\sec^2\theta}{(2\tan\theta)2\sqrt{\sec^2\theta}}d\theta\\&=\int_{\pi/6}^{\pi/4}\frac{\sec\theta}{2\tan\theta}d\theta=\frac12\int_{\pi/6}^{\pi/4}\frac{\cos\theta}{\sin\theta}\frac1{\cos\theta}d\theta\\&=\frac12\int_{\pi/6}^{\pi/4}\frac1{\sin\theta}d\theta=\frac12\int_{\pi/6}^{\pi/4}\csc\theta d\theta\\&=-\frac12[\ln|\csc\theta+\cot\theta|]_{\pi/6}^{\pi/4}\\&=-\frac12[\ln|\csc(\pi/4)+\cot(\pi/4)|-\ln|\csc(\pi/6)+\cot(\pi/6)|]\\&=-\frac12\ln\frac{\sqrt2+1}{2+\sqrt3}.\end{aligned}$$

(b)(i)

$x=A(1-x)+B$. By inspection $A=-1,B=1$.

$$\int\frac x{\sqrt{1-x}}dx=\int\frac{-(1-x)+1}{\sqrt{1-x}}dx=\int-(1-x)^{1/2}dx+\int(1-x)^{-1/2}dx=-\frac{(1-x)^{3/2}}{(3/2)(-1)}+\frac{(1-x)^{1/2}}{(-1)(1/2)}+C=\frac23(1-x)^{3/2}-2(1-x)^{1/2}+C\quad\text{(proven)}.$$

(b)(ii)

$\int\sqrt x dx=\frac23x^{3/2}+c$, and $d(\sin^{-1}\sqrt x)/dx=(1/\sqrt{1-x})(1/(2\sqrt x))$.

$$\begin{aligned}\int\sqrt x(\sin^{-1}\sqrt x)dx&=\frac23x^{3/2}\sin^{-1}\sqrt x-\int\frac23x^{3/2}\frac1{2\sqrt x}\frac1{\sqrt{1-x}}dx\\&=\frac23x^{3/2}\sin^{-1}\sqrt x-\frac13\int\frac x{\sqrt{1-x}}dx\\&=\frac23x^{3/2}\sin^{-1}\sqrt x-\frac13[\frac23(1-x)^{3/2}-2(1-x)^{1/2}+C]\\&=\frac23x^{3/2}\sin^{-1}\sqrt x-\frac29(1-x)^{3/2}+\frac23(1-x)^{1/2}+D.\end{aligned}$$

Teacher notes

Substitution: $\int f(x)dx=\int g(\theta)(dx/d\theta)d\theta$. Change limits from $x$ to $\theta$ for the definite integral; express everything in $\theta$. Note the printed warning uses $g(u)$ with $d\theta$: $\int g(u)(dx/d\theta)d\theta\ne(\int g(u)d\theta)(dx/d\theta)$.

MF27 $\int\csc x dx=-\ln|\csc x+\cot x|+c$; remember the modulus sign, replace $x$ by $\theta$.

(b)(i) uses the power formula. (b)(ii) uses parts with $u=\sin^{-1}\sqrt x$, $dv/dx=\sqrt x$.

Printed answer key

(a) $-\frac12\ln[(\sqrt2+1)/(2+\sqrt3)]$. (b)(i) $A=-1,B=1$; (ii) $\frac23x^{3/2}\sin^{-1}\sqrt x-\frac29(1-x)^{3/2}+\frac23(1-x)^{1/2}+D$.