king's property (1 Viewer)

tito981 · Oct 23, 2021

Is King's property assumed knowledge or are they more likely to tell you to derive it before using?

Trebla · Oct 23, 2021

If something is not explicitly called out in the syllabus, then obviously it is not assumed knowledge.

icycledough · Oct 23, 2021

tito981 said:
Is King's property assumed knowledge or are they more likely to tell you to derive it before using?

Not going to lie, having done Extension 2 last year, I've never come across the King's property nor have I seen it in writing anywhere (whether in a textbook, exam or in the syllabus)

CM_Tutor · Oct 24, 2021

The King property has appeared in MX2 and 4u Maths exams (HSCs and Trials) over many years. I don't recall a case where it was needed in an HSC without a suitable introduction, so I certainly would not take it as assumed knowledge... though it is a technique of which all MX2 students should be aware. I also can't recall having seen it mentioned in an HSC context as the "King property". In any case, it is the property that

$\int_a^b f(x)\ dx = \int_a^b f(a + b - x)\ dx$

that is usually introduced with a $u$ substitution. It is perhaps best known for questions like

$I = \int_0^{\frac{\pi}{2}} \cfrac{\sin^n{x}}{\sin^n{x}+\cos^n{x}}\ dx \qquad \qquad \text{. . . . . . . . . . (1)}$

where it allows us to show that

$I = \int_0^{\frac{\pi}{2}} \cfrac{\cos^n{x}}{\sin^n{x}+\cos^n{x}}\ dx \qquad \qquad \text{. . . . . . . . . . (2)}$

and thus gives the simplification, on adding equations (1) and (2)

$\begin{align*} I + I &= \int_0^{\frac{\pi}{2}} \cfrac{\sin^n{x}}{\sin^n{x}+\cos^n{x}}\ dx + \int_0^{\frac{\pi}{2}} \cfrac{\cos^n{x}}{\sin^n{x}+\cos^n{x}}\ dx \\ 2I &= \int_0^{\frac{\pi}{2}} \cfrac{\sin^n{x}+\cos^n{x}}{\sin^n{x}+\cos^n{x}}\ dx \\ &= \int_0^{\frac{\pi}{2}} 1\ dx \\ &= \bigg[ x \bigg]_0^{\frac{\pi}{2}} \\ &= \cfrac{\pi}{2} - 0 \\ I &= \cfrac{\pi}{4} \end{align*}$

It can be applied to solving integrals like

$\int_0^{\frac{\pi}{2}} \cfrac{dx}{1+(\tan{x})^\pi}$

$\int_0^{\frac{\pi}{2}} \cfrac{dx}{1+(\tan{x})^\pi} = \cfrac{\pi}{4}$

and

$\int_0^1 \cfrac{\ln{(x+1)}}{x^2 + 1}\ dx$

$\int_0^1 \cfrac{\ln{(x+1)}}{x^2 + 1}\ dx = \cfrac{\pi\ln{2}}{8}$

mathsbrain · Nov 6, 2021

CM_Tutor said:
The King property has appeared in MX2 and 4u Maths exams (HSCs and Trials) over many years. I don't recall a case where it was needed in an HSC without a suitable introduction, so I certainly would not take it as assumed knowledge... though it is a technique of which all MX2 students should be aware. I also can't recall having seen it mentioned in an HSC context as the "King property". In any case, it is the property that

$\int_a^b f(x)\ dx = \int_a^b f(a + b - x)\ dx$

that is usually introduced with a $u$ substitution. It is perhaps best known for questions like

$I = \int_0^{\frac{\pi}{2}} \cfrac{\sin^n{x}}{\sin^n{x}+\cos^n{x}}\ dx \qquad \qquad \text{. . . . . . . . . . (1)}$

where it allows us to show that

$I = \int_0^{\frac{\pi}{2}} \cfrac{\cos^n{x}}{\sin^n{x}+\cos^n{x}}\ dx \qquad \qquad \text{. . . . . . . . . . (2)}$

and thus gives the simplification, on adding equations (1) and (2)

$\begin{align*} I + I &= \int_0^{\frac{\pi}{2}} \cfrac{\sin^n{x}}{\sin^n{x}+\cos^n{x}}\ dx + \int_0^{\frac{\pi}{2}} \cfrac{\cos^n{x}}{\sin^n{x}+\cos^n{x}}\ dx \\ 2I &= \int_0^{\frac{\pi}{2}} \cfrac{\sin^n{x}+\cos^n{x}}{\sin^n{x}+\cos^n{x}}\ dx \\ &= \int_0^{\frac{\pi}{2}} 1\ dx \\ &= \bigg[ x \bigg]_0^{\frac{\pi}{2}} \\ &= \cfrac{\pi}{2} - 0 \\ I &= \cfrac{\pi}{4} \end{align*}$

It can be applied to solving integrals like

$\int_0^{\frac{\pi}{2}} \cfrac{dx}{1+(\tan{x})^\pi}$

$\int_0^{\frac{\pi}{2}} \cfrac{dx}{1+(\tan{x})^\pi} = \cfrac{\pi}{4}$

and

$\int_0^1 \cfrac{\ln{(x+1)}}{x^2 + 1}\ dx$

$\int_0^1 \cfrac{\ln{(x+1)}}{x^2 + 1}\ dx = \cfrac{\pi\ln{2}}{8}$

how do we do the last two?

5uckerberg · Nov 6, 2021

mathsbrain said:
how do we do the last two?

For this one $I=\int_{0}^{\frac{\pi}{2}}\frac{dx}{1+(\tan{x})^{\pi}}$
We will first let $u=\frac{\pi}{2}-x\boldsymbol{\Rightarrow}du=-dx$
Next step change the variables and there it becomes $\int_{\frac{\pi}{2}}^{0}\frac{-du}{1+(\tan({\frac{\pi}{2}-u}))^{\pi}}\boldsymbol{\Rightarrow}\int_{0}^{\frac{\pi}{2}}\frac{du}{1+(\tan({\frac{\pi}{2}-u}))^{\pi}}\boldsymbol{\Rightarrow}\int_{0}^{\frac{\pi}{2}}\frac{du}{1+(\cot{u})^{\pi}}$
The next step is the fun step because we will be doing here is this. We want $(\tan{u})^{\pi})$ . In turn, to retrieve $(\tan{u})^{\pi})$ from $\int_{0}^{\frac{\pi}{2}}\frac{du}{1+(\cot{u})^{\pi}}$ is to multiply $\int_{0}^{\frac{\pi}{2}}\frac{du}{1+(\cot{u})^{\pi}}$ by $(\tan{u})^{\pi})$ and there we have $\int_{0}^{\frac{\pi}{2}}\frac{(\tan{u})^{\pi}du}{(\tan{u})^{\pi}+1}=J$ noting that u is the dummy variable. Notice anything here, J=1-I
therefore, we have $2I=\int_{0}^\frac{\pi}{2}dx\boldsymbol{\Rightarrow}I=\frac{\pi}{4}$

5uckerberg · Nov 6, 2021

I think the first one could be in the HSC because it comes from the syllabus. However the second one I really think a little bit of hand-holding is needed to complete the question. Otherwise you will get nowhere.

CM_Tutor · Nov 6, 2021

5uckerberg said:
I think the first one could be in the HSC because it comes from the syllabus. However the second one I really think a little bit of hand-holding is needed to complete the question. Otherwise you will get nowhere.

Fair point. If this was an HSC question, it would likely tell you to use $x=\tan\theta$ before the King property.

5uckerberg · Nov 6, 2021

CM_Tutor said:
Fair point. If this was an HSC question, it would likely tell you to use $x=\tan\theta$ before the King property.

Ah yes, it then becomes $\int_{0}^{\frac{\pi}{4}}\ln{(1+\tan{\theta})}d\theta$ . The rest is textbook work

stupid_girl · Nov 7, 2021

Does this one require a little bit of hand-holding?

$\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx$

$\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx$
$=\frac{1}{2}\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)+\ln\left(\frac{1}{4}+\frac{3\tan\left(\frac{\pi}{10}+\frac{3\pi}{20}-x\right)-1}{1-3\tan^{2}\left(\frac{\pi}{10}+\frac{3\pi}{20}-x\right)}\right)dx$
$=\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\frac{3}{4}dx$
$=\frac{\pi}{20}\ln3-\frac{\pi}{10}\ln2$

By the way, there are variants that I often find more useful.
$\int_{b}^{a}f\left(x\right)dx=\int_{b}^{\frac{a+b}{2}}f\left(x\right)+f\left(a+b-x\right)dx=\int_{\frac{a+b}{2}}^{a}f\left(x\right)+f\left(a+b-x\right)dx$
Two common cases are b=-a and b=0.
$\int_{-a}^{a}f\left(x\right)dx=\int_{0}^{a}f\left(x\right)+f\left(-x\right)dx$
$\int_{0}^{a}f\left(x\right)dx=\int_{0}^{\frac{a}{2}}f\left(x\right)+f\left(a-x\right)dx$

One more example for fun:
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin x\cos x\ln\left(1+e^{x}\right)}{\sin^{4}x+\cos^{4}x}dx$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin x\cos x\ln\left(1+e^{x}\right)}{\sin^{4}x+\cos^{4}x}dx$
$=\int_{0}^{\frac{\pi}{2}}\left(\ln\left(1+e^{x}\right)-\ln\left(1+e^{-x}\right)\right)\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$=\int_{0}^{\frac{\pi}{2}}x\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$=\int_{0}^{\frac{\pi}{4}}\left(x+\frac{\pi}{2}-x\right)\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$=\frac{\pi}{2}\int_{0}^{\frac{\pi}{4}}\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$=\frac{\pi}{4}\int_{0}^{\frac{\pi}{4}}\frac{2\tan x\sec^{2}x}{1+\left(\tan^{2}x\right)^{2}}dx$
$=\frac{\pi}{4}\int_{0}^{1}\frac{1}{1+u^{2}}du$
$=\frac{\pi^{2}}{16}$

5uckerberg · Nov 7, 2021

stupid_girl said:
Does this one require a little bit of hand-holding?
$\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx$

$\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx$
$=\frac{1}{2}\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)+\ln\left(\frac{1}{4}+\frac{3\tan\left(\frac{\pi}{10}+\frac{3\pi}{20}-x\right)-1}{1-3\tan^{2}\left(\frac{\pi}{10}+\frac{3\pi}{20}-x\right)}\right)dx$
$=\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\frac{3}{4}dx$
$=\frac{\pi}{20}\ln3-\frac{\pi}{10}\ln2$

By the way, there are variants that I often find more useful.
$\int_{b}^{a}f\left(x\right)dx=\int_{b}^{\frac{a+b}{2}}f\left(x\right)+f\left(a+b-x\right)dx=\int_{\frac{a+b}{2}}^{a}f\left(x\right)+f\left(a+b-x\right)dx$
Two common cases are b=-a and b=0.
$\int_{-a}^{a}f\left(x\right)dx=\int_{0}^{a}f\left(x\right)+f\left(-x\right)dx$
$\int_{0}^{a}f\left(x\right)dx=\int_{0}^{\frac{a}{2}}f\left(x\right)+f\left(a-x\right)dx$

One more example for fun:
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin x\cos x\ln\left(1+e^{x}\right)}{\sin^{4}x+\cos^{4}x}dx$

Now, suppose

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin x\cos x\ln\left(1+e^{x}\right)}{\sin^{4}x+\cos^{4}x}dx$
$=\int_{0}^{\frac{\pi}{2}}\left(\ln\left(1+e^{x}\right)-\ln\left(1+e^{-x}\right)\right)\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$=\int_{0}^{\frac{\pi}{2}}x\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$=\int_{0}^{\frac{\pi}{4}}\left(x+\frac{\pi}{2}-x\right)\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$=\frac{\pi}{2}\int_{0}^{\frac{\pi}{4}}\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$=\frac{\pi}{4}\int_{0}^{\frac{\pi}{4}}\frac{2\tan x\sec^{2}x}{1+\left(\tan^{2}x\right)^{2}}dx$
$=\frac{\pi}{4}\int_{0}^{1}\frac{1}{1+u^{2}}du$
$=\frac{\pi^{2}}{16}$

Actually do you mean $\int_{b}^{a}f\left(x\right)dx=\int_{b}^{a}f\left(x\right)+f\left(a+b-x\right)dx=\int_{b}^{a}f\left(x\right)+f\left(a+b-x\right)dx$ . If that is the case then surely it exists and is the general form of the King Property. I reckon a little introduction might help and the rest is just what they were taught.

king's property (1 Viewer)

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