Solved Examples Using the Properties

This block of fifteen fully solved examples shows the properties of Section 4.2.1 combined with ordinary integration techniques (rationalising a surd denominator, half-angle/product-to-sum trigonometric identities, integration by parts inside a definite integral, partial fractions, and the greatest-integer function) — the same toolbox the Exercise 4.2 questions draw on.

Ex. 1. $\int_1^3\dfrac{1}{\sqrt{2+x}+\sqrt x}\,dx$: rationalise by multiplying by $\dfrac{\sqrt{2+x}-\sqrt x}{\sqrt{2+x}-\sqrt x}$; the denominator becomes $(2+x)-x=2$, so the integral is $\frac12\int_1^3(\sqrt{2+x}-\sqrt x)\,dx=\frac13\Big[(2+x)^{3/2}-x^{3/2}\Big]_1^3=\frac13\big\{5^{3/2}-2\cdot3^{3/2}+1\big\}$.

Ex. 2. $\int_0^{\pi/2}\sqrt{1-\cos4x}\,dx$: using $1-\cos A=2\sin^2(A/2)$ with $A=4x$ gives $\sqrt{2}\,|\sin2x|=\sqrt2\sin2x$ on $[0,\pi/2]$, so the integral is $\sqrt2\Big[-\frac{\cos2x}2\Big]_0^{\pi/2}=-\frac{\sqrt2}2(\cos\pi-\cos0)=-\frac{\sqrt2}2(-2)=\sqrt2$.

Ex. 3. $\int_0^{\pi/2}\cos^3x\,dx$: using $\cos^3x=\frac14(\cos3x+3\cos x)$ and integrating termwise gives $\frac14\Big[\frac{\sin3x}3+3\sin x\Big]_0^{\pi/2}=\frac14\big(-\frac13+3\big)=\frac14\cdot\frac83=\frac23$. (The text also shows the same value $2/3$ reached by the alternative substitution $t=\tan x$, reducing the integrand to a rational function of $t$ and finishing with a logarithmic partial-fraction integral — both routes must agree since they evaluate the same integral.)

Ex. 5. $\int_1^2\dfrac{\log x}{x^2}\,dx$: integrate by parts with $u=\log x$, $dv=x^{-2}dx$ so $v=-1/x$: $\Big[-\frac{\log x}{x}\Big]_1^2+\int_1^2\frac{1}{x^2}\,dx=\Big(-\frac{\log2}2-0\Big)+\Big[-\frac1x\Big]_1^2=-\frac{\log2}2+\Big(-\frac12+1\Big)=\frac12\big(1-\log2\big)$.

Ex. 6. $\int_0^{\pi/2}\dfrac{\cos x}{1+\cos x+\sin x}\,dx$: writing everything in terms of the half-angle $x/2$, the numerator and denominator both factor to reveal $\dfrac{\cos(x/2)-\sin(x/2)}{\cos(x/2)}=1-\tan(x/2)$; integrating gives $\Big[x-2\log\sec\frac x2\Big]_0^{\pi/2}=\frac\pi4-\log\sqrt2$.

Ex. 7. $\int_0^{1/2}\dfrac{1}{(1-2x^2)\sqrt{1-x^2}}\,dx$: substitute $x=\sin\theta$; the interval $[0,\tfrac12]$ becomes $[0,\pi/6]$, and $1-2\sin^2\theta=\cos2\theta$, $\sqrt{1-\sin^2\theta}=\cos\theta$, so the integrand becomes $\sec2\theta$; integrating and evaluating gives $\frac12\log(2+\sqrt3)$.

Ex. 8. $\int_0^2\dfrac{2^x}{2^x(1+4^x)}\,dx$: put $t=2^x$; after simplifying by partial fractions in $t$, the integral reduces to $\dfrac{1}{\log2}\log\!\Big(\dfrac{4\sqrt2}{\sqrt{17}}\Big)$.

Ex. 9 (absolute value). $\int_{-1}^1|5x-3|\,dx$: since $5x-3$ changes sign at $x=3/5$ (inside $[-1,1]$), split into $\int_{-1}^{3/5}-(5x-3)\,dx+\int_{3/5}^1(5x-3)\,dx$; evaluating both pieces and adding gives $\frac{34}5$.

Ex. 10 (a Property-V pairing template). $\int_0^{\pi/2}\dfrac{1}{1+\sqrt3\tan x}\,dx$: write $I$ for this integral; reflecting $x\to\frac\pi2-x$ turns $\tan x$ into $\cot x$ and gives a second expression for $I$ with sine and cosine swapped; adding the two expressions for $I$ makes the denominators identical, collapsing $2I$ to $\int_0^{\pi/2}1\,dx=\frac\pi2$, so $I=\frac\pi4$. This exact reflect-and-add pattern (call the integral $I$, apply $\int_a^bf=\int_a^bf(a+b-x)$, and add the two copies so the awkward denominator cancels) is the template behind several Exercise 4.2 Part II and III questions.

Ex. 11 (the general "$f/(f+f(a+b-x))$" pattern). $\int_3^8\dfrac{(11-x)^2}{x^2+(11-x)^2}\,dx$: reflecting $x\to3+8-x=11-x$ swaps the numerator and one part of the denominator with the other, so adding the original and reflected copies gives $2I=\int_3^8 1\,dx=5$, i.e. $I=\frac52$. In general, for any $f$, $\displaystyle\int_a^b\dfrac{f(x)}{f(x)+f(a+b-x)}\,dx=\frac12(b-a)$ — the same reflect-and-add trick as Ex. 10, stated as a reusable rule.

Ex. 12. $\int_0^\pi x\sin^2x\,dx$: reflecting $x\to\pi-x$ (using $\sin(\pi-x)=\sin x$) gives $I=\int_0^\pi(\pi-x)\sin^2x\,dx=\pi\int_0^\pi\sin^2x\,dx-I$, so $2I=\pi\int_0^\pi\frac12(1-\cos2x)\,dx=\frac\pi2\big[x-\tfrac12\sin2x\big]_0^\pi=\frac\pi2\cdot\pi=\frac{\pi^2}2$, giving $I=\frac{\pi^2}4$. …