Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

jamonwindeyer · « **Reply #210 on:** May 15, 2016, 02:04:12 am »

Quote from: amandali on May 15, 2016, 12:17:37 am

(Image removed from quote.)

im stuck after making sin^n(x) to sin^n-2(x)sin^2(x)

Hey Amandali! Your approach might work, but I think it will be easier to change that split a little, try splitting into sinx and sin^(n-1)x and doing integration by parts:

$u=\sin^{n-1}{x}\\ u'=(n-1)\sin^{n-2}{x}\cos{x}\\ v=-\cos{x}\\ v'=\sin{x}$

$I_n = [-\sin^{n-1}{x}\cos{x}]^\frac{\pi}{2}_0 + (n-1)\int^\frac{\pi}{2}_0\sin^{n-2}{x}\cos^2{x}dx$

Check the sign changes and substitutions in that second term to make sure you follow what I did. That first term will disappear, it evaluates to zero! So, we manipulate the second term:

$I_n = (n-1)\int^\frac{\pi}{2}_0\sin^{n-2}{x}(1-\sin^2{x})dx \\ =(n-1)(I_{n-2}-I_n)$
$I_n = nI_{n-2}-nI_n-I_{n-2}+I_n \\ nI_n=(n-1)I_{n-2}\\ I_n=\frac{n-1}{n}I_{n-2}$

Let me know if anything here was a little unclear! I skipped a little bit of algebra

hope it helps!

RuiAce · « **Reply #211 on:** May 15, 2016, 09:53:52 am »

Quote from: amandali on May 15, 2016, 12:17:37 am

(Image removed from quote.)

im stuck after making sin^n(x) to sin^n-2(x)sin^2(x)

$\text{You should not split up by a power of two for the sine and cosine power reduction integrals}$

$\textit{Obviously it can be integrated, but the integrals of }\sin^2{x}\text{ and }\cos^2{x}\textit{ are not tidy.}\\ \text{This is contrary to how it is tidy for }\sec^2{x}\\ \text{ which is why you do that for }\int{\sec^n{x}dx}$

I recommend that for the classic examples such as sinⁿ(x) you simply memorise the natural approach, which is to split it up by a power of one and not two in this case. (Method of course demonstrated by Jamon already.)

jakesilove · « **Reply #212 on:** May 15, 2016, 10:16:23 am »

Quote from: RuiAce on May 15, 2016, 09:53:52 am

$\text{You should not split up by a power of two for the sine and cosine power reduction integrals}$

$\textit{Obviously it can be integrated, but the integrals of }\sin^2{x}\text{ and }\cos^2{x}\textit{ are not tidy.}\\ \text{This is contrary to how it is tidy for }\sec^2{x}\\ \text{ which is why you do that for }\int{\sec^n{x}dx}$

I recommend that for the classic examples such as sinⁿ(x) you simply memorise the natural approach, which is to split it up by a power of one and not two in this case. (Method of course demonstrated by Jamon already.)

One day, I'm going to get to some Maths and Physics questions before you or Jamon.... One day!

amandali · « **Reply #213 on:** May 17, 2016, 12:31:25 am »

got stuck after using integration by parts

RuiAce · « **Reply #214 on:** May 17, 2016, 09:48:17 am »

Quote from: amandali on May 17, 2016, 12:31:25 am

(Image removed from quote.)

got stuck after using integration by parts

$\text{The first time you attempt this question you may find extreme difficulty}\\ \text{But once you see the plus-minus one trick, it will become easy.}$

$\begin{align*}{I}_{n}&=\int_{0}^{x}{{\left(1+{t}^{2}\right)}^{n}dt}\\&={\left[t{\left(1+{t}^{2}\right)}^{n}\right]}_{0}^{x}-\int_{0}^{x}{{t\cdot2nt\left(1+{t}^{2}\right)}^{n-1}dt}\\&=x{\left(1+{x}^{2}\right)}^{n}+0-2n\int_{0}^{x}{{{t}^{2}\left(1+{t}^{2}\right)}^{n-1}dt}\\&\stackrel{\pm 1}{=}x{\left(1+{x}^{2}\right)}^{n}-2n\int_{0}^{x}{{{\left(\left(1+{t}^{2}\right)-1\right)}^{2}\left(1+{t}^{2}\right)}^{n-1}dt}\\&=x{\left(1+{x}^{2}\right)}^{n}-2n\int_{0}^{x}{\left({\left(1+{t}^{2}\right)\left(1+{t}^{2}\right)}^{n-1}-\left(1+{t}^{2}\right)\right)dt}\\&=x{\left(1+{x}^{2}\right)}^{n}-2n\int_{0}^{x}{{\left(1+{t}^{2}\right)}^{n}dt}+2n\int_{0}^{x}{{\left(1+{t}^{2}\right)}^{n-1}dt}\\&=x{\left(1+{x}^{2}\right)}^{n}-2n{I}_{n}+2n{I}_{n-1}\\ \Leftrightarrow (2n+1)I_n&=x{\left(1+{x}^{2}\right)}^{n}+2n{I}_{n-1}\\ \Leftrightarrow I_n&=\frac{{\left(1+x^2\right)}^n}{2n+1}x+\frac{2n}{2n+1}I_{n-1}\end{align*}$

amandali · « **Reply #215 on:** May 17, 2016, 07:55:00 pm »

i did part (i) but not sure how it is used for part (ii)

RuiAce · « **Reply #216 on:** May 17, 2016, 08:56:12 pm »

Quote from: amandali on May 17, 2016, 07:55:00 pm

(Image removed from quote.)

i did part (i) but not sure how it is used for part (ii)

$\begin{align*}\frac{d}{dx}\left[x^{n-1}{\left(1+x^{2}\right)}^{\frac{3}{2}}\right]&=\left[(n+2)x^{n}+(n-1)x^{n-2}\right]\sqrt{1+x^{2}}\\\Leftrightarrow\frac{d}{dx}\left[x^{n-1}{\left(1+x^{2}\right)}^{\frac{3}{2}}\right]&=(n+2)x^{n}\sqrt{1+x^{2}}+(n-1)x^{n-2}\sqrt{1+x^{2}}\\\Rightarrow\int_{0}^{t}{\frac{d}{dx}\left[x^{n-1}{\left(1+x^{2}\right)}^{\frac{3}{2}}\right]\,dx}&=\int_{0}^{t}{\left[(n+2)x^{n}\sqrt{1+x^{2}}+(n-1)x^{n-2}\sqrt{1+x^{2}}\right]dx}\\\Leftrightarrow t^{n-1}{\left(1+t^{2}\right)}^{\frac{3}{2}}-0^{n-1}{\left(1+0^{2}\right)}^{\frac{3}{2}}&=(n+2)\int_{0}^{t}{x^{n}\sqrt{1+x^{2}}dx}+(n-1)\int_{0}^{t}{x^{n-2}\sqrt{1+x^{2}}dx}\\\Leftrightarrow t^{n-1}{\left(1+t^{2}\right)}^{\frac{3}{2}}&=(n+2)I_{n}+(n-1)I_{n-2}\\\Leftrightarrow(n+2)I_{n}&=t^{n-1}{\left(1+t^{2}\right)}^{\frac{3}{2}}-(n-1)I_{n-2}\end{align*}$

amandali · « **Reply #217 on:** May 18, 2016, 11:53:05 pm »

not sure how to do part b) part a is written on top

jakesilove · « **Reply #218 on:** May 19, 2016, 09:18:48 am »

Quote from: amandali on May 18, 2016, 11:53:05 pm

(Image removed from quote.)

not sure how to do part b) part a is written on top

Hey!

I've posted my solution below. This one was largely intuitive; once you've written out a couple expansions using your answer to a), it should be pretty obvious that b) is true.

Jake

RuiAce · « **Reply #219 on:** May 19, 2016, 01:06:26 pm »

Quote from: amandali on May 18, 2016, 11:53:05 pm

(Image removed from quote.)

not sure how to do part b) part a is written on top

Judging by your writing to the side it would appear that you tried to work from bottom to top. If you wanted to do this approach instead then the bad idea was the choice to simplify the expression.

$\begin{align*}I_0&=\frac{\pi}{2}\\I_2&=\frac{1}{2}I_0\\&=\frac{1}{2}\times \frac{\pi}{2}\\I_4&=\frac{3}{4}I_2\\&=\frac{3\times 1}{4\times 2}\times \frac{\pi}{2}\\I_6&=\frac{5}{6}I_4\\&=\frac{5\times3\times1}{6\times2\times4}\times\frac{\pi}{2}\\ &\vdots\\I_n&=\frac{(n-1)(n-3)\dots 3\times1}{(n-2)(n-4)\dots 4\times 2}\times\frac{\pi}{2}\end{align*}$

amandali · « **Reply #220 on:** May 23, 2016, 07:57:09 pm »

help with this ques thanks

RuiAce · « **Reply #221 on:** May 23, 2016, 09:14:28 pm »

Quote from: amandali on May 23, 2016, 07:57:09 pm

(Image removed from quote.)

help with this ques thanks

$\text{I actually had to experiment. This is the notorious }f(a-x)\text{ trick.}$
$\int_0^{\frac{\pi}{4}}{\frac{1-\sin{2x}}{1+\sin{2x}}dx}\\ =\int_0^{\frac{\pi}{4}}{\frac{1-\sin{2\left(\frac{\pi}{4}-x\right)}}{1+\sin{2\left(\frac{\pi}{4}-x\right)}}dx}\\=\int_0^{\frac{\pi}{4}}{\frac{1-\cos{2x}}{1+\cos{2x}}dx}\\=\int_0^{\frac{\pi}{4}}{\frac{2\sin^2{x}}{2\cos^2{x}}dx}=\int_0^\frac{\pi}{4}{\tan^2{x}dx}$

$\text{If you actually had to evaluate it you'd just swap }\tan^2{x}\text{ with }\sec^2{x}-1$

katherine123 · « **Reply #222 on:** May 27, 2016, 12:20:19 am »

what r the shortcut to simplify these

jamonwindeyer · « **Reply #223 on:** May 27, 2016, 11:51:02 am »

Quote from: katherine123 on May 27, 2016, 12:20:19 am

what r the shortcut to simplify these

Hey Katherine! Do you mean what steps they used to get to the end of each line?

I'll pull the binomial-style term out of each one to show you the process. I'll just do the outline, I might skip a step here or there, let me know if anything doesn't make sense!

Here are the first two:

$(a\sec{\theta}-a\cos{\theta})^2=a^2(\sec{\theta}-\cos{\theta})^2=\frac{a^2(1-\cos^2{\theta})^2}{\cos^2{\theta}}=\frac{a^2\sin^4{\theta}}{\cos^2{\theta}}=a^2\cos^2{\theta}\tan^4{\theta}$

$(b\tan{\theta}+b\cot{\theta})^2=\frac{b^2(\sin^2{\theta}+\cos^2{\theta})^2}{\sin^2{\theta}\cos^2{\theta}}=\frac{b^2}{\sin^2{\theta}\cos^2{\theta}}=b^2\csc^2{\theta}\sec^2{\theta}$

jamonwindeyer · « **Reply #224 on:** May 27, 2016, 12:03:26 pm »

Quote from: katherine123 on May 27, 2016, 12:20:19 am

what r the shortcut to simplify these

And here is the first term of the second two. Get a common term, factor out a sine on the top and a cosine on the bottom, you are left with -1 squared

$(a\sec{\theta}-\frac{a\cos{\theta}}{1-\sin{\theta}})^2 = a^2\left(\frac{1-\sin{\theta}-\cos^2{\theta}}{\cos{\theta}-\cos{\theta}\sin{\theta}}\right)^2=a^2\left(\frac{\sin^2{\theta}-\sin{\theta}}{\cos{\theta}-\sin{\theta}\cos{\theta}}\right)^2=a^2\left(\frac{\sin{\theta}}{\cos{\theta}}\right)^2\left(\frac{\sin{\theta}-1}{1-\sin{\theta}}\right)^2 = a^2\tan^2{\theta}$

The second term of the second two will be very similar. Try it! Put everything over a common term, and you'll be able to simplify the numerator with a pythagorean identity. Then, you'll notice that there there will be cancellation with the numerator and denominator, after you factor out a cosine on the bottom. The result should follow right after

hope this helped!

Search the forums now!

We have moved!

Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

jamonwindeyer

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

jakesilove

Re: 4U Maths Question Thread

amandali

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

amandali

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

amandali

Re: 4U Maths Question Thread

jakesilove

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

amandali

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

katherine123

Re: 4U Maths Question Thread

jamonwindeyer

Re: 4U Maths Question Thread

jamonwindeyer

Re: 4U Maths Question Thread

Recent Posts

User:
Password: