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[MagmaMcFry]

It's always a good idea to make more variables than less. Create a variable for every value that could possibly be relevant, then write down all elementary relations and use those to solve for the variable you need to solve for.

[ed boy]

I'm not having a good time with A2 maths (C3, C4, M2) at the moment

Would you mind being slightly more specific? I assume you came here to get help, right?

I'm just generally stuck on understanding some of the stuff and was wondering if there was something I could do to help me understand some of the concepts, rather than just practicing stuff I don't understand then forgetting it?

There is stuff you can do to help you understand some of the concepts. Either you simply Google for those concepts, or you tell us the names of the concepts (as well as some example problems using it) and we'll try our best to help you understand it.

I've been trying to go through my textbooks/class notes and using a maths site so far but I'm still not 100% there. I'll keep going through all the content we've covered and try some more, and I'll come back and ask if there is still stuff I don't fully understand.

Cheerd

It's been a few years since I did those modules, but I can help out if needs be.

[Vector]

It's always a good idea to make more variables than less. Create a variable for every value that could possibly be relevant, then write down all elementary relations and use those to solve for the variable you need to solve for.

Yeah, this one.

[Pnx]

So I'm trig integration, and I've got the problem of trying to find the arc length from 0 to pi/4 of the function y = ln(cos(x)).

I turned this into the integral of (1+sec²(x))^(1/2) But I'm really struggling to figure out what the heck I do with this. My book claims the answer is ln(2^(1/2) + 1).

[da_nang]

So I'm trig integration, and I've got the problem of trying to find the arc length from 0 to pi/4 of the function y = ln(cos(x)).

I turned this into the integral of (1+sec²(x))^(1/2) But I'm really struggling to figure out what the heck I do with this. My book claims the answer is ln(2^(1/2) + 1).

Check your derivative. I think it's wrong.

[Nightscar982]

I have to integrate trig soon. Not looking forward to it :L

[Skyrunner]

Here's a problem that was super difficult for me D:
Please tell me if it's difficult in  in y'all's eyes!

f(x) is a differentiable function with the domain 0 <x <pi/2.
For all valid x, f'(x) = cos^2 (f (x)), and f(1)=pi/4. if g (x) is the inverse function of f (x), find g(pi/3) * g'(pi/3).

Assume f'(x) means 'the first order differential of f(x)'.

[MagmaMcFry]

Let h = cosē. Then f'=h°f. Now 1 = (f°g)' = f'°g*g' = h°f°g*g' = h*g', therefore g' = 1/h = 1/cosē, and g = tan+c. Since g(pi/4) = 1, g = tan. Therefore g(pi/3)*g'(pi/3) = 4*sqrt(3).

Sounds hard enough to me, at least it wouldn't be a question you'd find in an exam.

[Skyrunner]

* Skyrunner claps

The problem, of course, lies in the fact that half of my math test problems are of comparable difficulty... 

...

[Pnx]

So I'm trig integration, and I've got the problem of trying to find the arc length from 0 to pi/4 of the function y = ln(cos(x)).

I turned this into the integral of (1+sec²(x))^(1/2) But I'm really struggling to figure out what the heck I do with this. My book claims the answer is ln(2^(1/2) + 1).

Check your derivative. I think it's wrong.

Bleh, silly me, thanks. I'm sure I'll stop making these mistakes someday.

[Pnx]

The integral of sin⁴(x)

It's embarrassing, I should really know how to do this one, but I'm sort of stumped. :|

[MagmaMcFry]

It helps if you write sin⁴(x) as ((exp(ix)-exp(-ix))/2i)^4 = (exp(4ix)-4*exp(2ix)+6-4*exp(-2ix)+exp(-4ix))/16.

Then integrating gives you (exp(4ix)/4i-4*exp(2ix)/2i+6x+4*exp(-2ix)/2i-exp(-4ix)/4i)/16 = 1/32*sin(4x)-1/4*sin(2x)+3/8x.

[Pnx]

I... do not understand what you just wrote at all.

[Skyrunner]

The integral of sin⁴(x)

It's embarrassing, I should really know how to do this one, but I'm sort of stumped. :|

sin⁴(x) = (1 - cos²(x))²=1-2cos ² (x) + cos⁴(x)
Turn cos²(x) into ( 1 + cos (0.5x))/2, and turn cos⁴(x) into (1+cos²(0.5x))/2, then into  (1+( (1+cos (0.25))/2)/2. Now it's integratable!

[MagmaMcFry]

sin⁴(x) = (1 - cos²(x))²=1-2cos ² (x) + cos⁴(x)
Turn cos²(x) into ( 1 + cos (0.5x))/2, and turn cos⁴(x) into (1+cos²(0.5x))/2, then into  (1+( (1+cos (0.25))/2)/2. Now it's integratable!

First, cosē(x) = (1+cos(2x))/2, not (1+cos(0.5x))/2; also cos⁴(x) = ((1+cos(2x))/2)^2, not (1+cosē(2x))/2. But the idea still works, you eventually find out that cos⁴(x) = 1/8*cos(4x) - 1/2*cos(2x) + 3/8.