Need help please (1 Viewer)

hihihi · Jul 7, 2016

bump - for my questoin, its been 6 min not 5 integrand-ationgrandfatherclock

but srsly, i need help cuz im going to sleep soon

InteGrand · Jul 7, 2016

hihihi said:
also need halp with this q:

π
cos^6(x) x sin(x) dx
0

upper bound is pi and lower is 0

im getting answer as zero, i kinda need explantion on how bounds change when using integrating but substiintuition

$\noindent You should write `*' for the multiplication sign, otherwise it looks like an $x$. Anyway, for that integral, let $u = \cos x$, so $\mathrm{d}u =-\sin x \text{ d}x$. When $x = 0, u = \cos 0 = 1$, and when $x = \pi$, $u = \cos \pi = -1$. So $\int _0 ^{\pi} \cos^6 x \sin x \text{ d}x = \int _{-1}^{1} u^6 \text{ d}u = \frac{2}{7}$.$

hihihi · Jul 7, 2016

ok tyvsmsmvm

so ur just supposed to sub in the u values for a bound? aye see

how about if its like

let u = 3x^5 * 2

are u supposed so say the bound is now

3u^5 *2

or like wut?

hihihi · Jul 7, 2016

btw does anyone or anytwo are anythree know what marking u got ghetto out of 60 to get hd in olympiadomatic?

InteGrand · Jul 7, 2016

hihihi said:
ok tyvsmsmvm

so ur just supposed to sub in the u values for a bound? aye see

how about if its like

let u = 3x^5 * 2

are u supposed so say the bound is now

3u^5 *2

or like wut?

For the bounds you essentially replace them with the value of the new variable when the old variable equals those bounds. Like if the bounds were 0 to 16 and we did a substitution of u = sqrt(x) (where the original variable was x), then the bounds for u would be from 0 to 4, since when x = 0, u = 0, and when x = 16, u = 4.

hihihi · Jul 7, 2016

oohhh okay

would it be 5u maths if it was like a bound of -1 in a circleroot(x) question? like the one u stated above?

i mean squearetoot

leehuan · Jul 7, 2016

leehuan said:
No need to bump after just six minutes

$$This is just $ \\ \begin{align*}V&=\pi \int_{\frac{1}{2}}^1 \cot^2{\frac{\pi}{2}x}\,dx\\ &= \pi \int_{\frac{1}{2}}^1 \left(\csc^2{\frac{\pi}{2}x}-1 \right)\end{align*}$

$\text{Note that integrating cosec-squared gives cot}$

hihihi said:
soz if the bump affected u

can u solve it fully,

cuz as i implied, i got a different asnwer, so like, is it okie dokie if u can do it all for me

i know its 11:20 and u probs got idk uni or school in 5 mins but like, pl0x it wont take long

thx

p.s.s.s.s.s great u good man XDXDXD

Depends on what you mean by affect. It didn't bother me, just some courtesy

Ok so if I evaluate the integral out

$\begin{align*}V&=\pi \left[ -\frac{2}{\pi}\cot \left(\frac{\pi}{2}x \right) - x\right]_\frac{1}{2}^1\\ &= \pi \left( -\frac{2}{\pi} \cot \frac{\pi}{2} - 1 + \frac{2}{\pi}\cot \frac{\pi}{4} + \frac{1}{2} \right) \\ &=\pi \left( \frac{2}{\pi}-\frac{1}{2} \right) \\ &= \frac{1}{2}(4-\pi) \end{align*}$

InteGrand · Jul 7, 2016

hihihi said:
oohhh okay

would it be 5u maths if it was like a bound of -1 in a circleroot(x) question? like the one u stated above?

i mean squearetoot

Yeah

Nailgun · Jul 7, 2016

leehuan said:
Depends on what you mean by affect. It didn't bother me, just some courtesy

Ok so if I evaluate the integral out

$\begin{align*}V&=\pi \left[ -\frac{2}{\pi}\cot \left(\frac{\pi}{2}x \right) - x\right]_\frac{1}{2}^1\\ &= \pi \left( -\cot \frac{\pi}{2} - 1 + \cot \frac{\pi}{4} + \frac{1}{2} \right) \\ &=\pi \left( \frac{2}{\pi}-\frac{1}{2} \right) \\ &= \frac{1}{2}(4-\pi) \end{align*}$

I did not know that you can integrate cosec^2(x) into -cot(x)

lol should I know that?

InteGrand · Jul 7, 2016

Nailgun said:
I did not know that you can integrate cosec^2(x) into -cot(x)

lol should I know that?

Derivative of sec(x) is sec(x)tan(x), derivative cot(x) is -csc^2 (x) and derivative of csc(x) is -csc(x)cot(x).

Reversing these relations gives us some antiderivatives.

These derivatives are listed in the Year 12 3U Pender (Cambridge) textbook I believe (because they're essentially derivatives of standard HSC functions), but I'm not sure whether you'd need to know them off the top of your head for 2U (even in 3U/4U, getting asked about them seems to be on the rarer side).

hihihi · Jul 7, 2016

affect1
əˈfɛkt/
verb
have an effect on; make a difference to.
"the dampness began to affect my health"
synonyms: influence, exert influence on, have an effect on, act on, work on, condition, touch, have an impact on, impact on, take hold of, attack, infect, strike, strike at, hit; More

as stated by the consituion of bristish law upon the act of the commonweatlthy games of 1991 under king henry the 8th.

btw, im a bit confused where that 2/pi went in the second line... and isnt cot(pi/2) = 0 cuz like tan(pi/2) is undefined?

thx

leehuan · Jul 7, 2016

Nailgun said:
I did not know that you can integrate cosec^2(x) into -cot(x)

lol should I know that?

It's similar to how sec²(x) integrates into tan(x)

It may be worth memorising the reciprocal trigonometric function derivatives. (I'll leave the proof as an exercise, but it's just using the chain rule.)

$\frac{d}{dx}\sec x = \sec x \tan x\\ \frac{d}{dx}\cot x = -\csc^2 x \\ \frac{d}{dx} \csc x = -\csc x \cot x$

Haven't seen them appear in 2U out of the blue though. Sec has appeared, but they said use your table of standard integrals.

leehuan · Jul 7, 2016

hihihi said:
affect1
əˈfɛkt/
verb
have an effect on; make a difference to.
"the dampness began to affect my health"
synonyms: influence, exert influence on, have an effect on, act on, work on, condition, touch, have an impact on, impact on, take hold of, attack, infect, strike, strike at, hit; More

as stated by the consituion of bristish law upon the act of the commonweatlthy games of 1991 under king henry the 8th.

btw, im a bit confused where that 2/pi went in the second line... and isnt cot(pi/2) = 0 cuz like tan(pi/2) is undefined?

thx

That 2/π disappearing was my mistake - I'll fix it back in there.

And yes of course cot(π/2) = 0

hihihi · Jul 7, 2016

gotcha thx homie from da underground below sealevel yea im in love with the coco

hihihi · Jul 7, 2016

btw can someone answer that olypmipaid question;;

it was like:

what mark do i ghetto get out of 60 to get the high distinction

like i know its a percentage or something

but loike, around what mark to be in toppish 10%

tyvm boiz catcha all tomorrow boiz g'nite y'all from underground

and ill see my fans later y'all *coco*

Nailgun · Jul 7, 2016

InteGrand said:
Derivative of sec(x) is sec(x)tan(x), derivative cot(x) is -csc^2 (x) and derivative of csc(x) is -csc(x)cot(x).

Reversing these relations gives us some antiderivatives.

These derivatives are listed in the Year 12 3U Pender (Cambridge) textbook I believe (because they're essentially derivatives of standard HSC functions), but I'm not sure whether you'd need to know them off the top of your head for 2U (even in 3U/4U, getting asked about them seems to be on the rarer side).

leehuan said:
It's similar to how sec²(x) integrates into tan(x)

It may be worth memorising the reciprocal trigonometric function derivatives. (I'll leave the proof as an exercise, but it's just using the chain rule.)

$\frac{d}{dx}\sec x = \sec x \tan x\\ \frac{d}{dx}\cot x = -\csc^2 x \\ \frac{d}{dx} \csc x = -\csc x \cot x$

Haven't seen them appear in 2U out of the blue though. Sec has appeared, but they said use your table of standard integrals.

Tenks. Sense maketh.

Sec has appeared, but they said use your table of standard integrals.

Ah, I think I did that question. I had to pause the exam and go download the old formula sheet. The new one doesn't have it I'm pree sure.

Green Yoda · Jul 8, 2016

Nailgun said:
You have an uh...interesting way of typing.

Azn confirmed

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