Maths-Ext 1/2 Marks Lost and Hardest Topics (2 Viewers)

CM_Tutor · Jan 19, 2021

vernburn said:
Here is how $x=\cos2\theta$ sub works out:
--- remove equation that is not displaying ---

Now I'm sure you can see how this works out from here - far from elegant but it still works!

As for another trig sub, maybe $x=\tan^2\theta$ :
--- remove equation that is not displaying ---

Neither of those equations is displaying for me. However, $x=\tan^2\theta$ is the trig substitution that I know does works, and so I have modified your post and elaborated a bit so that it does display, as follows, for those who want to see the working:

$I=\int{\frac{x^6}{(1+x)^8}}\ dx$
$=\int{\frac{\tan^{12}\theta}{(1+\tan^2\theta)^8}}\times2\tan{\theta}\sec^2\theta\ d\theta \qquad \text{by taking $x=\tan^2\theta \implies dx=2\tan{\theta}\sec^2\theta\ d\theta$}$
$=\int{\frac{2\tan^{13}\theta\sec^2\theta}{\sec^{16}\theta}}\ d\theta \qquad \text{using the Pythagorean identity $\tan^2\theta + 1 = \sec^2\theta$}$
$=2\int{\frac{\sin^{13}\theta}{\cos^{13}\theta}\div\frac{1}{\cos^{14}\theta}}\ d\theta \qquad \text{as $\sec{\theta}=\frac{1}{\cos{\theta}}$}$
$=2\int{\sin^{13}\theta\times\cos{\theta}}\ d\theta$
$= 2\int{u^{13}}\ du \qquad \text{where $u=\sin{\theta} \implies du=\cos{\theta}\ d\theta$}$
$=\frac{2u^{14}}{14}+C \qquad \text{for some constant $C$}\qquad \text{. . . . . . . (*)}$
$=\left(\frac{\sqrt{x}}{\sqrt{x+1}}\right)^{14}\times\frac{1}{7}+C \qquad \text{as $u=\sin{\theta}=\sin{\left(\tan^{-1}\sqrt{x}\right)}=\frac{\sqrt{x}}{\sqrt{x+1}}$}$
$=\frac{x^7}{7(1+x)^7}+C$
$\begin{align*} \text{Or: by noting that (*) can be re-written as:} \quad I&=\frac{2u^{14}}{14}+C=\frac{1}{7}\left(u^2\right)^7+C \\ \text{and $u^2$ can then be calculated as follows:} \quad \sec^2\theta&=\tan^2\theta+1 \quad \text{and $x$ was defined as} \quad x=\tan^2\theta \\ \frac{1}{\cos^2\theta}&=x+1 \\ \cos^2\theta&=\frac{1}{x+1} \\ 1-\sin^2\theta&=\frac{1}{x+1} \\ \sin^2{\theta}&= 1-\frac{1}{x+1} \\ u^2 &=\frac{x}{x+1} \\ \text{So,} \quad I&=\frac{1}{7}\times \left(\frac{x}{x+1}\right)^7+C \\ &=\frac{x^7}{7(1+x)^7}+C \end{align*}$

I don't have time to look at the other solution at this moment, but will check it later today.

Qeru · Jan 19, 2021

idkkdi said:
anyone bothered to finish off the sectan integrating in the cos2theta substitution lol.? btw, u would factor out sec tan , and yeet that out with a sec theta = u substitution.

any smarter way than doing that algebra in cos2theta solution?

Yes but really difficult to spot is to use the 'reverse quotient rule,' whenever we are integrating a rational function it;s always a good idea to keep this trick in mind.

First note that $7x^6(1+x)^7-7x^7(1+x)^6=7x^6(1+x)^6(1+x-x)=7x^6(1+x)^6$
So rewriting the integral:

$I=\frac{1}{7}\int{\frac{7x^6(1+x)^7-7x^7(1+x)^6}{(1+x)^{14}}}\ dx$

This is in the form $\int{\frac{u'v-uv'}{v^2} dx$ where $u=x^7 \quad \text{and} \quad v=(1+x)^7$

So $I=\frac{x^7}{7(1+x)^7}+C$

idkkdi · Jan 19, 2021

Jojofelyx said:
my Cambridge ext 1 is like 750 lmao :\\

But isn't getting a band 6 based of the difficulty, I read something on SMH that was about extension subjects and having higher % of kids who get band 6, mainly cuz its something they are good at and willing to invest time into it, that's why music extension got like 50% or something band 6s?idk could be wrong, feel free to correct me tho.

and wdym algebra grind? copious amounts of working out or wot?

cambridge ext 1 for yr 12's even longer iirc and 4u questions are probably another level up from what you do in 3u.

idrc about this band 6 stuff.

some tricky working out. specific things you need to do to get to an answer.
And yes, copious amounts of working out in certain cases.

Jojofelyx · Jan 19, 2021

idkkdi said:
cambridge ext 1 for yr 12's even longer iirc and 4u questions are probably another level up from what you do in 3u.

idrc about this band 6 stuff.

some tricky working out. specific things you need to do to get to an answer.
And yes, copious amounts of working out in certain cases.

Ah shit, I see. anyway too good for band 6's, they better make a b7 just for you xo

Directrix · Jan 20, 2021

@Qeru and @idkkdi are the origins of the “2 smartest kids in the classroom” meme

idkkdi · Jan 20, 2021

Jojofelyx said:
Ah shit, I see. anyway too good for band 6's, they better make a b7 just for you xo

band 6 is like a 75 raw. anyone like @Qeru is aiming for 95 raw lmao.

CM_Tutor · Jan 20, 2021

I said I would post looking at the $x=\cos2\theta$ approach. @vernburn was correct, it does work... but it's certainly not as easy a solution as the $x=\tan^2{\theta}$ approach, and to show that it matches the standard result that

$I=\int\frac{x^6}{(1+x)^8}\ dx = \frac{x^7}{7(1+x)^7}+C \quad \text{for some constant $C$}$

requires some difficult binomial work (especially to complete without any structure provided). So, here it is:

$I=\int\frac{x^6}{(1+x)^8}\ dx$

$=\int{\frac{\cos^6{2\theta}}{(1+\cos{2\theta})^8}} \times -2\sin2\theta\ d\theta \quad \text{by taking $x=\cos{2\theta} \implies dx=-\sin{2\theta}\times2$}$

$=-2\int\frac{\left(2\cos^2\theta-1\right)^6}{(2\cos^2\theta)^8}\times2\sin\theta\cos\theta d\theta \quad \text{using the identities $\cos2\theta=2\cos^2\theta-1$ and $\sin2\theta=2\sin\theta\cos\theta$}$

$=\frac{-2\times2}{2^8}\int{\frac{\left(2\cos^2\theta-1\right)^6\sin\theta\cos\theta}{\cos^{16}\theta}}\ d\theta$

$=\frac{-1}{2^6}\int{\frac{\left(2^6\binom{6}{0}\cos^{12}\theta-2^5\binom{6}{1}\cos^{10}\theta+2^4\binom{6}{2}\cos^8\theta-2^3\binom{6}{3}\cos^6\theta+2^2\binom{6}{4}\cos^4\theta-2\binom{6}{5}\cos^2\theta+\binom{6}{6}\right)\sin\theta}{\cos^{15}\theta}}\ d\theta$

$=\frac{-1}{2^6}\int\left(\frac{2^6\binom{6}{0}}{\cos^3\theta}-\frac{2^5\binom{6}{1}}{\cos^5\theta}+\frac{2^4\binom{6}{2}}{\cos^7\theta}-\frac{2^3\binom{6}{3}}{\cos^9\theta}+\frac{2^2\binom{6}{4}}{\cos^{11}\theta}-\frac{2\binom{6}{5}}{\cos^{13}\theta}+\frac{\binom{6}{6}}{\cos^{15}\theta}\right)\sin\theta\ d\theta$

$=\frac{1}{2^6}\int{\frac{2^6\binom{6}{0}}{u^3}-\frac{2^5\binom{6}{1}}{u^5}+\frac{2^4\binom{6}{2}}{u^7}-\frac{2^3\binom{6}{3}}{u^9}+\frac{2^2\binom{6}{4}}{u^{11}}-\frac{2\binom{6}{5}}{u^{13}}+\frac{\binom{6}{6}}{u^{15}}}\ du \quad \text{by taking $u=\cos\theta \implies du=-\sin\theta$}$

$=\frac{1}{2^6}\left(\frac{2^6\binom{6}{0}}{-2u^2}-\frac{2^5\binom{6}{1}}{-4u^4}+\frac{2^4\binom{6}{2}}{-6u^6}-\frac{2^3\binom{6}{3}}{-8u^8}+\frac{2^2\binom{6}{4}}{-10u^{10}}-\frac{2\binom{6}{5}}{-12u^{12}}+\frac{\binom{6}{6}}{-14u^{14}}\right)+C \quad \text{for some constant $C$}$

$=\frac{1}{2^6}\left(-\frac{2^5\binom{6}{0}}{u^2}+\frac{2^3\binom{6}{1}}{u^4}-\frac{2^3\binom{6}{2}}{3u^6}+\frac{\binom{6}{3}}{u^8}-\frac{2\binom{6}{4}}{5u^{10}}+\frac{\binom{6}{5}}{2\times3u^{12}}-\frac{\binom{6}{6}}{2\times7u^{14}}\right)+C$

$=-\frac{\binom{6}{0}}{2u^2}+\frac{\binom{6}{1}}{2^3u^4}-\frac{\binom{6}{2}}{2^3\times3u^6}+\frac{\binom{6}{3}}{2^6u^8}-\frac{\binom{6}{4}}{2^5\times5u^{10}}+\frac{\binom{6}{5}}{2^7\times3u^{12}}-\frac{\binom{6}{6}}{2^7\times7u^{14}}+C$

$=\frac{-1}{u^2}\left(\frac{\binom{6}{0}}{2}-\frac{\binom{6}{1}}{2^3u^2}+\frac{\binom{6}{2}}{2^3\times3u^4}-\frac{\binom{6}{3}}{2^6u^6}+\frac{\binom{6}{4}}{2^5\times5u^8}-\frac{\binom{6}{5}}{2^7\times3u^{10}}+\frac{\binom{6}{6}}{2^7\times7u^{12}}\right)+C$

$=\frac{-1}{2u^2}\left[\binom{6}{0}+\frac{1}{2}\binom{6}{1}\left(\frac{-1}{2u^2}\right)+\frac{1}{3}\binom{6}{2}\left(\frac{-1}{2u^2}\right)^2+\frac{1}{4}\binom{6}{3}\left(\frac{-1}{2u^2}\right)^3+\frac{1}{5}\binom{6}{4}\left(\frac{-1}{2u^2}\right)^4+\frac{1}{6}\binom{6}{5}\left(\frac{-1}{2u^2}\right)^5+\frac{1}{7}\binom{6}{6}\left(\frac{-1}{2u^2}\right)^6\right]+C$

$=\frac{-1}{2u^2}\left[\binom{6}{0}+\frac{1}{2}\binom{6}{1}\left(\frac{-1}{2u^2}\right)+\frac{1}{3}\binom{6}{2}\left(\frac{-1}{2u^2}\right)^2+\frac{1}{4}\binom{6}{3}\left(\frac{-1}{2u^2}\right)^3+\frac{1}{5}\binom{6}{4}\left(\frac{-1}{2u^2}\right)^4+\frac{1}{6}\binom{6}{5}\left(\frac{-1}{2u^2}\right)^5+\frac{1}{7}\binom{6}{6}\left(\frac{-1}{2u^2}\right)^6\right]+C \quad \text{. . . . . . . (*)}$

Now, consider that:

$(1+z)^6 = \binom{6}{0} + \binom{6}{1}z + \binom{6}{2}z^2 + \binom{6}{3}z^3 + \binom{6}{4}z^4 + \binom{6}{5}z^5 + \binom{6}{6}z^6$

$\int{(1+z)^6}\ dz = \int{\binom{6}{0} + \binom{6}{1}z + \binom{6}{2}z^2 + \binom{6}{3}z^3 + \binom{6}{4}z^4 + \binom{6}{5}z^5 + \binom{6}{6}z^6}\ dz$

$\frac{(1+z)^7}{7} = \binom{6}{0}z + \binom{6}{1}\frac{z^2}{2} + \binom{6}{2}\frac{z^3}{3} + \binom{6}{3}\frac{z^4}{4} + \binom{6}{4}\frac{z^5}{5} + \binom{6}{5}\frac{z^6}{6} + \binom{6}{6}\frac{z^7}{7}+D \quad \text{for some constant $D$}$

$\text{Put $z=0$:}\quad \frac{1}{7} = 0+0+0+0+0+0+0+C \implies D=\frac{1}{7}$

$\frac{(1+z)^7 - 1}{7} = \binom{6}{0}z + \binom{6}{1}\frac{z^2}{2} + \binom{6}{2}\frac{z^3}{3} + \binom{6}{3}\frac{z^4}{4} + \binom{6}{4}\frac{z^5}{5} + \binom{6}{5}\frac{z^6}{6} + \binom{6}{6}\frac{z^7}{7}$

$\frac{(1+z)^7 - 1}{7z} = \binom{6}{0} + \binom{6}{1}\frac{z}{2} + \binom{6}{2}\frac{z^2}{3} + \binom{6}{3}\frac{z^3}{4} + \binom{6}{4}\frac{z^4}{5} + \binom{6}{5}\frac{z^5}{6} + \binom{6}{6}\frac{z^6}{7}$

$\text{Put } z = \frac{-1}{2u^2}: \\ \binom{6}{0}+\frac{1}{2}\binom{6}{1}\left(\frac{-1}{2u^2}\right)+\frac{1}{3}\binom{6}{2}\left(\frac{-1}{2u^2}\right)^2+\frac{1}{4}\binom{6}{3}\left(\frac{-1}{2u^2}\right)^3+\frac{1}{5}\binom{6}{4}\left(\frac{-1}{2u^2}\right)^4+\frac{1}{6}\binom{6}{5}\left(\frac{-1}{2u^2}\right)^5+\frac{1}{7}\binom{6}{6}\left(\frac{-1}{2u^2}\right)^6 \\ = \cfrac{\left(1+\cfrac{-1}{2u^2}\right)^7 - 1}{7\times\cfrac{-1}{2u^2}} \\ = \cfrac{\left(\cfrac{2u^2-1}{2u^2}\right)^7 - 1}{\cfrac{-7}{2u^2}} \times \cfrac{\left(2u^2\right)^7}{\left(2u^2\right)^7} \\ = \cfrac{\left(2u^2\right)^7 - \left(2u^2-1\right)^7}{7\left(2u^2\right)^6} \quad \text{. . . . . . . (**)}$

So, returning to the integral and putting (**) into (*), we have:

$I = \frac{-1}{2u^2} \times \cfrac{\left(2u^2\right)^7 - \left(2u^2-1\right)^7}{7\left(2u^2\right)^6} + C \quad \text{. . . . . . . (***)}$

And we have $u=\cos\theta$ where $\cos{2\theta} = x$ and so:

$\begin{align*} \cos{2\theta} &= 2\cos^2\theta - 1 \\ x &= 2u^2 - 1 \\ 2u^2 &= x+1 \end{align*}$

And so, from (***):

$I = \frac{-1}{x+1} \times \frac{(x+1)^7 - (x+1-1)^7}{7(x+1)^6} + C$

$= \frac{x^7 - (x+1)^7}{7(x+1)^7} + C$

$= \frac{x^7}{7(x+1)^7} - \frac{(x+1)^7}{7(x+1)^7} + C$

$= \frac{x^7}{7(x+1)^7} - \frac{1}{7} + C$

$= \frac{x^7}{7(x+1)^7} +C' \quad \text{where } C'=C-\frac{1}{7}$

So, the result is the same but with a different constant of integration.

Qeru · Jan 20, 2021

It didn't load for me as well but looking at what @idkkdi said it seems like we convert it to sec and tan, but then wouldn't $x=\cos{\theta}$ be better heres my working:

$dx=-\sin\theta d\theta$

$\therefore I=-\int{\frac{\cos^6{\theta}\sin\theta}{(1+\cos{\theta})^8} dx$

$I=-\int{\frac{\cos^7{\theta}.\frac{\sin\theta}{\cos\theta}}{(1+\cos{\theta})^8} dx$

$I=-\int{\frac{\frac{1}{\sec^7{\theta}}.\tan{\theta}}{(1+\frac{1}{\sec{\theta}})^8} dx$

$I=-\int{\frac{\sec\theta \tan\theta}{(1+\sec{\theta})^8} dx \quad \text{by multiplying top and bottom by} \sec^8{\theta}$

$I=\frac{1}{7(1+\sec\theta)^7}+C \quad \text{since}\quad \frac{d}{d\theta}(\sec{\theta})=\sec\theta\tan\theta$

$I=\frac{1}{7\left(1+\frac{1}{x}\right)^7}+C \quad \text{since}\quad x=\cos{\theta}=\frac{1}{\sec{\theta}} \implies \frac{1}{x}=\sec{\theta}$

$I=\frac{x^7}{7(1+x)^7}+C \quad \text{by multiplying top and bottom by} \quad x^7$

idkkdi · Jan 20, 2021

Qeru said:
It didn't load for me as well but looking at what @idkkdi said it seems like we convert it to sec and tan, but then wouldn't $x=\cos{\theta}$ be better heres my working:

$dx=-\sin\theta d\theta$

$\therefore I=-\int{\frac{\cos^6{\theta}\sin\theta}{(1+\cos{\theta})^8} dx$

$I=-\int{\frac{\cos^7{\theta}.\frac{\sin\theta}{\cos\theta}}{(1+\cos{\theta})^8} dx$

$I=-\int{\frac{\frac{1}{\sec^7{\theta}}.\tan{\theta}}{(1+\frac{1}{\sec{\theta}})^8} dx$

$I=-\int{\frac{\sec\theta \tan\theta}{(1+\sec{\theta})^8} dx \quad \text{by multiplying top and bottom by} \sec^8{\theta}$

$I=\frac{1}{7(1+\sec\theta)^7}+C \quad \text{since}\quad \frac{d}{d\theta}(\sec{\theta})=\sec\theta\tan\theta$

$I=\frac{1}{7\left(1+\frac{1}{x}\right)^7}+C \quad \text{since}\quad x=\cos{\theta}=\frac{1}{\sec{\theta}} \implies \frac{1}{x}=\sec{\theta}$

$I=\frac{x^7}{7(1+x)^7}+C \quad \text{by multiplying top and bottom by} \quad x^7$

CM_Tutor said:
I said I would post looking at the $x=\cos2\theta$ approach. @vernburn was correct, it does work... but it's certainly not as easy a solution as the $x=\tan^2{\theta}$ approach, and to show that it matches the standard result that

$I=\int\frac{x^6}{(1+x)^8}\ dx = \frac{x^7}{7(1+x)^7+C \quad \text{for some constant $C$}$

requires some difficult binomial work (especially to complete without any structure provided). So, here it is:

$I=\int\frac{x^6}{(1+x)^8}\ dx$

$=\int{\frac{\cos^6{2\theta}}{(1+\cos{2\theta})^8}} \times -2\sin2\theta\ d\theta \quad \text{by taking $x=\cos{2\theta} \implies dx=-\sin{2\theta}\times2$}$

$=-2\int\frac{\left(2\cos^2\theta-1\right)^6}{(2\cos^2\theta)^8}\times2\sin\theta\cos\theta d\theta \quad \text{using the identities $\cos2\theta=2\cos^2\theta-1$ and $\sin2\theta=2\sin\theta\cos\theta$}$

$=\frac{-2\times2}{2^8}\int{\frac{\left(2\cos^2\theta-1\right)^6\sin\theta\cos\theta}{\cos^{16}\theta}}\ d\theta$

$=\frac{-1}{2^6}\int{\frac{\left(2^6\binom{6}{0}\cos^{12}\theta-2^5\binom{6}{1}\cos^{10}\theta+2^4\binom{6}{2}\cos^8\theta-2^3\binom{6}{3}\cos^6\theta+2^2\binom{6}{4}\cos^4\theta-2\binom{6}{5}\cos^2\theta+\binom{6}{6}\right)\sin\theta}{\cos^{15}\theta}}\ d\theta$

$=\frac{-1}{2^6}\int\left(\frac{2^6\binom{6}{0}}{\cos^3\theta}-\frac{2^5\binom{6}{1}}{\cos^5\theta}+\frac{2^4\binom{6}{2}}{\cos^7\theta}-\frac{2^3\binom{6}{3}}{\cos^9\theta}+\frac{2^2\binom{6}{4}}{\cos^{11}\theta}-\frac{2\binom{6}{5}}{\cos^{13}\theta}+\frac{\binom{6}{6}}{\cos^{15}\theta}\right)\sin\theta\ d\theta$

$=\frac{1}{2^6}\int{\frac{2^6\binom{6}{0}}{u^3}-\frac{2^5\binom{6}{1}}{u^5}+\frac{2^4\binom{6}{2}}{u^7}-\frac{2^3\binom{6}{3}}{u^9}+\frac{2^2\binom{6}{4}}{u^{11}}-\frac{2\binom{6}{5}}{u^{13}}+\frac{\binom{6}{6}}{u^{15}}}\ du \quad \text{by taking $u=\cos\theta \implies du=-\sin\theta$}$

$=\frac{1}{2^6}\left(\frac{2^6\binom{6}{0}}{-2u^2}-\frac{2^5\binom{6}{1}}{-4u^4}+\frac{2^4\binom{6}{2}}{-6u^6}-\frac{2^3\binom{6}{3}}{-8u^8}+\frac{2^2\binom{6}{4}}{-10u^{10}}-\frac{2\binom{6}{5}}{-12u^{12}}+\frac{\binom{6}{6}}{-14u^{14}}\right)+C \quad \text{for some constant $C$}$

$=\frac{1}{2^6}\left(-\frac{2^5\binom{6}{0}}{u^2}+\frac{2^3\binom{6}{1}}{u^4}-\frac{2^3\binom{6}{2}}{3u^6}+\frac{\binom{6}{3}}{u^8}-\frac{2\binom{6}{4}}{5u^{10}}+\frac{\binom{6}{5}}{2\times3u^{12}}-\frac{\binom{6}{6}}{2\times7u^{14}}\right)+C$

$=-\frac{\binom{6}{0}}{2u^2}+\frac{\binom{6}{1}}{2^3u^4}-\frac{\binom{6}{2}}{2^3\times3u^6}+\frac{\binom{6}{3}}{2^6u^8}-\frac{\binom{6}{4}}{2^5\times5u^{10}}+\frac{\binom{6}{5}}{2^7\times3u^{12}}-\frac{\binom{6}{6}}{2^7\times7u^{14}}+C$

$=\frac{-1}{u^2}\left(\frac{\binom{6}{0}}{2}-\frac{\binom{6}{1}}{2^3u^2}+\frac{\binom{6}{2}}{2^3\times3u^4}-\frac{\binom{6}{3}}{2^6u^6}+\frac{\binom{6}{4}}{2^5\times5u^8}-\frac{\binom{6}{5}}{2^7\times3u^{10}}+\frac{\binom{6}{6}}{2^7\times7u^{12}}\right)+C$

$=\frac{-1}{2u^2}\left[\binom{6}{0}+\frac{1}{2}\binom{6}{1}\left(\frac{-1}{2u^2}\right)+\frac{1}{3}\binom{6}{2}\left(\frac{-1}{2u^2}\right)^2+\frac{1}{4}\binom{6}{3}\left(\frac{-1}{2u^2}\right)^3+\frac{1}{5}\binom{6}{4}\left(\frac{-1}{2u^2}\right)^4+\frac{1}{6}\binom{6}{5}\left(\frac{-1}{2u^2}\right)^5+\frac{1}{7}\binom{6}{6}\left(\frac{-1}{2u^2}\right)^6\right]+C$

$=\frac{-1}{2u^2}\left[\binom{6}{0}+\frac{1}{2}\binom{6}{1}\left(\frac{-1}{2u^2}\right)+\frac{1}{3}\binom{6}{2}\left(\frac{-1}{2u^2}\right)^2+\frac{1}{4}\binom{6}{3}\left(\frac{-1}{2u^2}\right)^3+\frac{1}{5}\binom{6}{4}\left(\frac{-1}{2u^2}\right)^4+\frac{1}{6}\binom{6}{5}\left(\frac{-1}{2u^2}\right)^5+\frac{1}{7}\binom{6}{6}\left(\frac{-1}{2u^2}\right)^6\right]+C \quad \text{. . . . . . . (*)}$

Now, consider that:

$(1+z)^6 = \binom{6}{0} + \binom{6}{1}z + \binom{6}{2}z^2 + \binom{6}{3}z^3 + \binom{6}{4}z^4 + \binom{6}{5}z^5 + \binom{6}{6}z^6$

$\int{(1+z)^6}\ dz = \int{\binom{6}{0} + \binom{6}{1}z + \binom{6}{2}z^2 + \binom{6}{3}z^3 + \binom{6}{4}z^4 + \binom{6}{5}z^5 + \binom{6}{6}z^6}\ dz$

$\frac{(1+z)^7}{7} = \binom{6}{0}z + \binom{6}{1}\frac{z^2}{2} + \binom{6}{2}\frac{z^3}{3} + \binom{6}{3}\frac{z^4}{4} + \binom{6}{4}\frac{z^5}{5} + \binom{6}{5}\frac{z^6}{6} + \binom{6}{6}\frac{z^7}{7}+D \quad \text{for some constant $D$}$

$\text{Put $z=0$:}\quad \frac{1}{7} = 0+0+0+0+0+0+0+C \implies D=\frac{1}{7}$

$\frac{(1+z)^7 - 1}{7} = \binom{6}{0}z + \binom{6}{1}\frac{z^2}{2} + \binom{6}{2}\frac{z^3}{3} + \binom{6}{3}\frac{z^4}{4} + \binom{6}{4}\frac{z^5}{5} + \binom{6}{5}\frac{z^6}{6} + \binom{6}{6}\frac{z^7}{7}$

$\frac{(1+z)^7 - 1}{7z} = \binom{6}{0} + \binom{6}{1}\frac{z}{2} + \binom{6}{2}\frac{z^2}{3} + \binom{6}{3}\frac{z^3}{4} + \binom{6}{4}\frac{z^4}{5} + \binom{6}{5}\frac{z^5}{6} + \binom{6}{6}\frac{z^6}{7}$

$\text{Put } z = \frac{-1}{2u^2}: \\ \binom{6}{0}+\frac{1}{2}\binom{6}{1}\left(\frac{-1}{2u^2}\right)+\frac{1}{3}\binom{6}{2}\left(\frac{-1}{2u^2}\right)^2+\frac{1}{4}\binom{6}{3}\left(\frac{-1}{2u^2}\right)^3+\frac{1}{5}\binom{6}{4}\left(\frac{-1}{2u^2}\right)^4+\frac{1}{6}\binom{6}{5}\left(\frac{-1}{2u^2}\right)^5+\frac{1}{7}\binom{6}{6}\left(\frac{-1}{2u^2}\right)^6 \\ = \cfrac{\left(1+\cfrac{-1}{2u^2}\right)^7 - 1}{7\times\cfrac{-1}{2u^2}} \\ = \cfrac{\left(\cfrac{2u^2-1}{2u^2}\right)^7 - 1}{\cfrac{-7}{2u^2}} \times \cfrac{\left(2u^2\right)^7}{\left(2u^2\right)^7} \\ = \cfrac{\left(2u^2\right)^7 - \left(2u^2-1\right)^7}{7\left(2u^2\right)^6} \quad \text{. . . . . . . (**)}$

So, returning to the integral and putting (**) into (*), we have:

$I = \frac{-1}{2u^2} \times \cfrac{\left(2u^2\right)^7 - \left(2u^2-1\right)^7}{7\left(2u^2\right)^6} + C \quad \text{. . . . . . . (***)}$

And we have $u=\cos\theta$ where $\cos{2\theta} = x$ and so:

$\begin{align*} \cos{2\theta} &= 2\cos^2\theta - 1 \\ x &= 2u^2 - 1 \\ 2u^2 &= x+1 \end{align*}$

And so, from (***):

$I = \frac{-1}{x+1} \times \frac{(x+1)^7 - (x+1-1)^7}{7(x+1)^6} + C$

$= \frac{x^7 - (x+1)^7}{7(x+1)^7} + C$

$= \frac{x^7}{7(x+1)^7} - \frac{(x+1)^7}{7(x+1)^7} + C$

$= \frac{x^7}{7(x+1)^7} - \frac{1}{7} + C$

$= \frac{x^7}{7(x+1)^7} +C' \quad \text{where } C'=C-\frac{1}{7}$

So, the result is the same but with a different constant of integration.

open image in new tab with right click for vernburn. it doesnt load for me either lol.
btw i meant at the u = costheta step sub in CM's solutions you could factor out a sectan and use a use a sec substitution.

Jojofelyx · Jan 20, 2021

idkkdi said:
band 6 is like a 75 raw. anyone like @Qeru is aiming for 95 raw lmao.

lmao its weird to think aa 75 raw gets around 90, and a 92 raw would get like a 95-96, 17 more marks raw = 5-6 more HSC marks, rip in peace.

quickoats · Jan 20, 2021

Jojofelyx said:
lmao its weird to think aa 75 raw gets around 90, and a 92 raw would get like a 95-96, 17 more marks raw = 5-6 more HSC marks, rip in peace.

Yep there’s definitely diminishing returns. 50 raw goes to like 80 whereas 99 raw goes to 99 and 100 raw goes to 100.

CM_Tutor · Jan 20, 2021

Qeru said:
It didn't load for me as well but looking at what @idkkdi said it seems like we convert it to sec and tan, but then wouldn't $x=\cos{\theta}$ be better heres my working:

$dx=-\sin\theta d\theta$

$\therefore I=-\int{\frac{\cos^6{\theta}\sin\theta}{(1+\cos{\theta})^8} dx$

$I=-\int{\frac{\cos^7{\theta}.\frac{\sin\theta}{\cos\theta}}{(1+\cos{\theta})^8} dx$

$I=-\int{\frac{\frac{1}{\sec^7{\theta}}.\tan{\theta}}{(1+\frac{1}{\sec{\theta}})^8} dx$

$I=-\int{\frac{\sec\theta \tan\theta}{(1+\sec{\theta})^8} dx \quad \text{by multiplying top and bottom by} \sec^8{\theta}$

$I=\frac{1}{7(1+\sec\theta)^7}+C \quad \text{since}\quad \frac{d}{d\theta}(\sec{\theta})=\sec\theta\tan\theta$

$I=\frac{1}{7\left(1+\frac{1}{x}\right)^7}+C \quad \text{since}\quad x=\cos{\theta}=\frac{1}{\sec{\theta}} \implies \frac{1}{x}=\sec{\theta}$

$I=\frac{x^7}{7(1+x)^7}+C \quad \text{by multiplying top and bottom by} \quad x^7$

Yep, there are multiple substitutions possible. I think the $x=\cos{2\theta}$ and then $u=\cos\theta$ method would actually be simpler as a direct $x=2u^2-1$ substitution.

The most efficient, though, is your reverse quotient rule approach. I have never seen that before in more than 20 years of tutoring etc, so I am very impressed!

vernburn · Jan 20, 2021

CM_Tutor said:
The most efficient, though, is your reverse quotient rule approach. I have never seen that before in more than 20 years of tutoring etc, so I am very impressed!

Look at the 2014 4U last question. Many people used the reverse quotient rule to get it out in a few lines lol

Qeru · Jan 20, 2021

vernburn said:
Look at the 2014 4U last question. Many people used the reverse quotient rule to get it out in a few lines lol

Yup thats where I got that technique from. For anyone interested, the integral is: $\int\frac{\ln x}{(1+\ln x)^2} dx$

Qeru · Jan 20, 2021

Also try this integral from the MIT integration bee:
$\int\sin^{99}(x)\sin(101x)dx$

Randomstudent123 · Jan 20, 2021

Qeru said:
Yup thats where I got that technique from. For anyone interested, the integral is: $\int\frac{\ln x}{(1+\ln x)^2} dx$

idk how to use latex so x/1+lnx +c

Randomstudent123 · Jan 20, 2021

Qeru said:
Also try this integral from the MIT integration bee:
$\int\sin^{101}(x)\sin(99x)dx$

isn't the MIT one sin(101x)sin^99x

Qeru · Jan 20, 2021

Randomstudent123 said:
isn't the MIT one sin(101x)sin^99x

My bad I remembered it wrong.

Randomstudent123 · Jan 20, 2021

Qeru said:
Also try this integral from the MIT integration bee:
$\int\sin^{99}(x)\sin(101x)dx$

then the answer is (sin^100x*sin(100x))/100

u expand sin(101x) and then sub u=sin^100xsin(100x)

Nacseo · Jan 20, 2021

Qeru said:
Yes but really difficult to spot is to use the 'reverse quotient rule,' whenever we are integrating a rational function it;s always a good idea to keep this trick in mind.

First note that $7x^6(1+x)^7-7x^7(1+x)^6=7x^6(1+x)^6(1+x-x)=7x^6(1+x)^6$
So rewriting the integral:

$I=\frac{1}{7}\int{\frac{7x^6(1+x)^7-7x^7(1+x)^6}{(1+x)^{14}}}\ dx$

This is in the form $\int{\frac{u'v-uv'}{v^2} dx$ where $u=x^7 \quad \text{and} \quad v=(1+x)^7$

So $I=\frac{x^7}{7(1+x)^7}+C$

Jeez my head hurt just by looking at these equations, I'm also gonna apply for Ext 1 Maths, but now when I see this I'm like, sh*t, SH*T.

Maths-Ext 1/2 Marks Lost and Hardest Topics (2 Viewers)

Moderator

Well-Known Member

Well-Known Member

Well-Known Member

Forklift Certified

Well-Known Member

Moderator

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Moderator

Active Member

Well-Known Member

Well-Known Member

New Member

New Member

Well-Known Member

New Member

Newest Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)