Author Topic: 4U Maths Question Thread (Read 665296 times) Tweet Share

jamonwindeyer · « **Reply #90 on:** March 22, 2016, 12:16:07 pm »

Quote from: RuiAce on March 22, 2016, 08:13:07 am

Actually, hang on (in regards to the diagram). Shouldn't the product of roots be negative s?

That's funny, there is also a mistake in the 3rd last line. It should be negative beta, so for the poster, the two incorrect lines:

$-\alpha^2\beta=-s$
$qr=(-\beta)(-\alpha^2)$

They happened to cancel each other out

amandali · « **Reply #91 on:** March 25, 2016, 12:03:55 am »

Is it better if i memorise this or show working out?

katherine123 · « **Reply #92 on:** March 25, 2016, 07:47:38 am »

Im stucked with this ques
What i did was changing sec^3(x) --> secx(1+tan^2(x)
so 1/cosx(1+sin^2(x)/cos^2(x))

katherine123 · « **Reply #93 on:** March 25, 2016, 08:23:32 am »

I dont get
-1/n[cosxsin^(n-1)x]
since I(n)= sin^n(x)

jamonwindeyer · « **Reply #94 on:** March 25, 2016, 12:27:03 pm »

Quote from: amandali on March 25, 2016, 12:03:55 am

Is it better if i memorise this or show working out?

Hey amandali!! It isn't in your table of standard integrals, so I think you best show working out in an exam. If it isn't in that table, you aren't expected to know it. However, it is definitely worth remembering as a security measure to check yourself as you go!!

jamonwindeyer · « **Reply #95 on:** March 25, 2016, 12:51:12 pm »

Quote from: katherine123 on March 25, 2016, 07:47:38 am

Im stucked with this ques
What i did was changing sec^3(x) --> secx(1+tan^2(x)
so 1/cosx(1+sin^2(x)/cos^2(x))

Hey Katherine! That is not the approach I would use, though it might work! I'll show how I would do it with integration by parts:

$I = \int sec^3(x)dx$

$u=\sec{x} \quad u'=\sec{x}\tan{x} \quad v=\tan{x} \quad v'=\sec^2{x}$

Through integration by parts:

$I = \sec{x}\tan{x} - \int \sec{x}tan^2{x}dx \\ = \sec{x}\tan{x}-\int(\sec^3{x}-\sec{x})dx \\ =\sec{x}\tan{x} - I + \int\sec{x}dx$

Now the integration of sec(x) is another job in itself. There is a common trick for figuring out this integral. Transform the integral like so:

$\int\sec{x}dx = \int\sec{x}\frac{\sec{x}+\tan{x}}{\sec{x}+\tan{x}}dx$

Then, use a substitution:

$u = \sec{x}+\tan{x}$

Follow this process to find that:

$\int\sec{x}dx=\ln|\sec{x}+\tan{x}|+C$

Taking the formula from above and adding this, and rearranging, will give us the final solution:

$I = \frac{1}{2}\sec{x}\tan{x}+\frac{1}{2}\ln|\sec{x}+\tan{x}|+C$

Let me know if anything here is unclear!! It is a long and difficult question

jamonwindeyer · « **Reply #96 on:** March 25, 2016, 01:04:26 pm »

Quote from: katherine123 on March 25, 2016, 08:23:32 am

I dont get
-1/n[cosxsin^(n-1)x]
since I(n)= sin^n(x)

Hey Katherine! We have another job for integration by parts here!

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

By integration by parts:

$I_n = \sin^{n-1}{x}\cos{x} - (n-1)\int\sin^{n-2}{x}\cos^2{x}dx \\ = \sin^{n-1}{x}\cos{x} - (n-1)\int\sin^{n-2}{x}-\sin^n{x}dx \\ = \sin^{n-1}{x}\cos{x} - (n-1)(I_{n-2}-I_n)$

Try rearranging this to obtain the solution given above:

$I_n=\frac{-1}{n}\sin^{n-1}{x}\cos{x}+\frac{n-1}{n}I_{n-2}$

I hope this helps!!

RuiAce · « **Reply #97 on:** March 25, 2016, 06:21:19 pm »

Quote from: amandali on March 25, 2016, 12:03:55 am

Is it better if i memorise this or show working out?

I'm sorry, but Jamon is right. You have to use partial fractions or a trigonometric substitution for that one.

Anyway side note @Jamon I'm sorry but please take care of any questions HPL can't! I'm not sure when Jake is back but I'm on a vacation at Melbourne right now with grandparents so I might miss the posts. Thanks!

jamonwindeyer · « **Reply #98 on:** March 25, 2016, 06:33:33 pm »

Quote from: RuiAce on March 25, 2016, 06:21:19 pm

Quote from: amandali on March 25, 2016, 12:03:55 am
Is it better if i memorise this or show working out?

I'm sorry, but Jamon is right. You have to use partial fractions or a trigonometric substitution for that one.

Anyway side note @Jamon I'm sorry but please take care of any questions HPL can't! I'm not sure when Jake is back but I'm on a vacation at Melbourne right now with grandparents so I might miss the posts. Thanks!

All good Rui! I'm on it, enjoy Melbourne and cheers for your help!

Happy Physics Land · « **Reply #99 on:** March 25, 2016, 08:18:34 pm »

Quote from: RuiAce on March 25, 2016, 06:21:19 pm

I'm sorry, but Jamon is right. You have to use partial fractions or a trigonometric substitution for that one.

Anyway side note @Jamon I'm sorry but please take care of any questions HPL can't! I'm not sure when Jake is back but I'm on a vacation at Melbourne right now with grandparents so I might miss the posts. Thanks!

Jake should be back tonight actually

jakesilove · « **Reply #100 on:** March 25, 2016, 08:33:30 pm »

Quote from: Happy Physics Land on March 25, 2016, 08:18:34 pm

Jake should be back tonight actually

HPL is right! I'm back and scrolling the forums, you guys have all done an incredible job keeping up to date on questions. I seriously, seriously appreciate the hard work you've all put in!

But yeah, back from my break and ready to answer some forum questions!

Thank guys; you are absolutely instrumental in maintaining the growth of this community, and we here at Atar Notes couldn't appreciate you more.

Jake

Happy Physics Land · « **Reply #101 on:** March 25, 2016, 09:22:50 pm »

Quote from: jakesilove on March 25, 2016, 08:33:30 pm

HPL is right! I'm back and scrolling the forums, you guys have all done an incredible job keeping up to date on questions. I seriously, seriously appreciate the hard work you've all put in!

But yeah, back from my break and ready to answer some forum questions!

Thank guys; you are absolutely instrumental in maintaining the growth of this community, and we here at Atar Notes couldn't appreciate you more.

Jake

Damn Jamon and Rui, we can't keep all the questions to ourselves anymore...
Jake just rocks up and breaks the equilibrium we have been trying to maintain for the past 2 weeks, and according to Le Chatelier's principle .... hehehe

RuiAce · « **Reply #102 on:** March 25, 2016, 10:35:53 pm »

Quote from: Happy Physics Land on March 25, 2016, 09:22:50 pm

Damn Jamon and Rui, we can't keep all the questions to ourselves anymore...
Jake just rocks up and breaks the equilibrium we have been trying to maintain for the past 2 weeks, and according to Le Chatelier's principle .... hehehe

Oh my goodness.

WRONG. THREAD.

jamonwindeyer · « **Reply #103 on:** March 25, 2016, 11:51:32 pm »

Quote from: Happy Physics Land on March 25, 2016, 09:22:50 pm

Damn Jamon and Rui, we can't keep all the questions to ourselves anymore...
Jake just rocks up and breaks the equilibrium we have been trying to maintain for the past 2 weeks, and according to Le Chatelier's principle .... hehehe

Hey come on man I didn't even do Chemistry! Don't confuse me

Happy Physics Land · « **Reply #104 on:** March 25, 2016, 11:54:34 pm »

Quote from: jamonwindeyer on March 25, 2016, 11:51:32 pm

Hey come on man I didn't even do Chemistry! Don't confuse me

Yeah I forgot you did legal studies, well in this CASE, I reckon you do have the right to remain silent ...

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