VCE Specialist 3/4 Question Thread!

[Bestie]

I know it's late and all but if someone has time, can you please also help me here...
if dy/dx = -1/[2x(squareroot x^2 - 1)]
and y = tan^-1[sqaureroot [(x+1)/(x-1)]]
then shouldn't the intergral of 1/(squareroot e^2x -1) just be -2e^2x[tan^-1[sqaureroot [(x+1)/(x-1)]]]
or is it becasue there is an x involved that i can't just simply take it out like a coefficient... how else would i find the intergral of 1/(squareroot e^2x -1), when i know that :
dy/dx = -1/[2x(squareroot x^2 - 1)]
and y = tan^-1[sqaureroot [(x+1)/(x-1)]]

thanks in advance. thank you so so much

[lzxnl]

What you do is multiply the top and bottom by e^x. Then, make a substitution u = e^x
You'll find that your integral should look like the integral you were given the answer to

[Bestie]

like this? (please see attachment)
but im stuck on what to do next?

[kinslayer]

Where did the -2 come from  (I hope I am looking at the right problem)

My first instinct would be to divide top and bottom by e^x, or equivalently factor the denominator like this:

(remember x > 0)

[Bestie]

yea.. you are looking at the right one but

the question is:
if dy/dx = -1/[2x(squareroot x^2 - 1)]
and y = tan^-1[sqaureroot [(x+1)/(x-1)]]
find that integral...
so the ans has like a tan^-1(.....) in it

[kinslayer]

OK, sorry, I misunderstood the question. Let's try this way:

[Bestie]

OMG yes thank you i got it now

if you have time can you please help me here:
how did they do this?
i was able to get the log bit but where did that boxed bit come from?

[kinslayer]

Partial fractions:



So:



*changed partial fraction calc slightly

[yang_dong]

Hi All!!!

I have a question: can someone please help?

A boat is being pulled into a wharf by a rope at a speed of 26 metres per minute.
If the rope is attached to a point on the boat 5 metres vertically below the wharf, at what rate is
the rope being drawn in, when the boat is 12 metres from the wharf?

thanks in advance

[Mieow]

Can someone please help me with these questions:
a)

b)

Thank You!

[b^3]

a) Hint: Make use of

b) Hint: Make use of

Spoiler

See if you can have a go at the second one now

EDIT: 3500^th post

[eagles]

Hi! Is there a way to integrate (log_ex)² by hand?

Thank you

[Homer]

integral of sin(x) is -cos(x)
integral of sin^3(x) is done differently

[lzxnl]

Hey, can anyone please check my working?  (ie. why is it wrong?)

   
Using cosine double angle formula, cos(2x) = 1 - 2sin^2(x)

Your flaw here is integrating with respect to sin x in the last line. What you've essentially done is let u = sin x and retain the dx in the integral, then mistakenly integrating u - 2u^3 dx with respect to u, even though it says dx.

I think you mean 1- 2sin^2 (x/2) in the second line
But really, you should have sin(x/2) (2cos^2(x/2)-1) instead on the second line to be able to substitute u = cos x/2. Note how we have a factor of sin(x/2) outside, which is a multiple of the derivative of cos x/2?

Hi! Is there a way to integrate (log_ex)² by hand?

Thank you

Differentiate x (ln x)^2 to get (ln x)^2 + 2 ln x
Now differentiate x ln x to get ln x + 1
Can you see that if we then differentiate x (ln x)^2 - 2 x ln x, we would be left with (ln x)^2 - 2?
See if you can do the rest

More methodically, you would use something called integration by parts, in which you reverse the product rule.
d(uv)/dx = u dv/dx + v du/dx
so uv = integral of u dv/dx dx + integral of v du/dx dx
or integral of u dv/dx dx = uv - integral of v du/dx dx
Note how instead of integrating the product of u and the derivative of v, we integrate the product of v and the derivative of u. So, when using this formula, we want to let u be something that simplifies upon differentiation and dv/dx be something not too complicated when integrated.
For integral of (ln x)^2, therefore, let u = (ln x)^2, dv/dx = 1, du/dx = 2 ln x / x, v = x
Formula gives integral of (ln x)^2 = x (ln x)^2 - int(2 ln x dx)
You can integrate ln x using u = ln x, dv/dx = 1

[eagles]

Awesome, thanks for showing the two methods.
But I have some questions about them that I was wondering if you can please clarify:

Differentiate x (ln x)^2 to get (ln x)^2 + 2 ln x
Now differentiate x ln x to get ln x + 1
Can you see that if we then differentiate x (ln x)^2 - 2 x ln x, we would be left with (ln x)^2 - 2?
See if you can do the rest
Yep, I understand the working. But how would we know which expressions to choose to differentiate?

More methodically, you would use something called integration by parts, in which you reverse the product rule.
d(uv)/dx = u dv/dx + v du/dx
so uv = integral of u dv/dx dx + integral of v du/dx dx
or integral of u dv/dx dx = uv - integral of v du/dx dx
Note how instead of integrating the product of u and the derivative of v, we integrate the product of v and the derivative of u. What do you mean by this? Aren't we integrating both values? Just that one is on the LHS of the equation and the other is on the RHS?
So, when using this formula, we want to let u be something that simplifies upon differentiation and dv/dx be something not too complicated when integrated.
For integral of (ln x)^2, therefore, let u = (ln x)^2, dv/dx = 1, du/dx = 2 ln x / x, v = x How is v = x?
Formula gives integral of (ln x)^2 = x (ln x)^2 - int(2 ln x dx)
You can integrate ln x using u = ln x, dv/dx = 1