4u Mathematics Marathon V 1.0 (1 Viewer)

[icycloud]

Oh god i did it a heaps long way lol. It involved 2cos@.

Oh well, i thought it was hard because no one responded soon after.

How did you do it your way? u = 2cos@? If that's the case, it's a similar approach to Keypad's. Oh and Keypad doesn't have a new question. Got another integration one? =D

 

[Riviet]

Sorry i meant cos2@ lol. This is how i designed this question:

∫6cos³xsinx dx=∫3cos²x2sinxcosx dx

=∫3cos²xsin2x dx

And the i let u=cos²x and changed it into 1/2(1+cos2x) to differentiate lol

 

[icycloud]

Wow, interesting alternative method I must say. Oh and your method yields the same result as Keypad's!

 

[Riviet]

Hey it's not that long i just realised lol, it's just that i took a while to make the question.

 

[KeypadSDM]

Hmm... that is alot faster than how i had to do it. Are you sure that's right keypad?

I hate being right. And no, I don't have any good questions. I just answer them

 

[icycloud]

Sorry i meant cos2@ lol. This is how i designed this question:

∫6cos³xsinx dx=∫3cos²x2sinxcosx dx

=∫3cos²xsin2x dx

And the i let u=cos²x and changed it into 1/2(1+cos2x) to differentiate lol

Oh BTW, instead of using u=cos²x, one could also do:

=3∫cos²xsin(2x) dx
=3∫(1/2 + cos(2x)/2) * sin(2x) dx
=3/2∫sin(2x)dx + 3/2∫cos(2x)sin(2x)dx
=-3/4 cos(2x) + 3/4∫sin(4x)dx
=-3/4 cos(2x) - 3/16 cos(4x) + C#
which differs by a constant to the other answer.

 

[Riviet]

Wow so many ways to do it.

Icycloud you can do the next question.

 

[icycloud]

Haha alright. I won't be replying till this arvo though, going to watch Domino@Fox now .

Next question:
∫Sqrt[(x-a)/(b-x)]dx

Looks easy, doesn't it? =D

 

[YBK]

everyone's soo ahead of me. we haven't done any integration, let alone 4unit integration!!!

(leaves me out of the marathon )

 

[Riviet]

Don't worry, integration is soo easy to learn and so much better than differentiating.

 

[KeypadSDM]

∫Sqrt[(x-a)/(b-x)]dx

If you pick a good substitution it comes out FAIRLY quickly. But not easily, mind you.

I'd feel better if I could just do it by parts [which I could, it's just a royal pain in the rear end, and for once I don't feel guilty by not doing it].

Spoiler

I = ∫Sqrt[(x-a)/(b-x)]dx
= ∫Sqrt[x-a]/Sqrt[b-x]dx
= ∫Sqrt[x-a]/Sqrt[(a - x) + (b - a)]dx

Now set u = Sqrt[x-a]
du = dx/(2Sqrt[x-a]) = dx/(2u)
dx = 2udu

Thus the integral becomes (noting that u² = x - a)

I = ∫u/Sqrt[-u² + (b - a)] * (2udu)
=2∫u²/Sqrt[-u² + (b - a)]du

Setting K² = b - a [For simplicity, we'll see how this unwinds ...]
And you get a nice integral to to by parts twice:

*Aside note: Why does b > a for this particular method to work?*

I = 2∫u²/Sqrt[K² - u²]du

Now we could do a nice substitution of u = Kcos[t] or some such, but I hate subs. Well let's delay it by 1 step in any case

Expanding the above we find:

I/2 = ∫[u² - K² + K²]/Sqrt[K² - u²]du
=∫[u² - K²]/Sqrt[K² - u²]du + ∫K²/Sqrt[K² - u²]du

Note that the second sum is a standard integral, evaluating to:
K² * Sin^-1[u/K]

Thus we have [by subtraction]:
I/2 - K² * Sin^-1[u/K]
= - ∫Sqrt[K² - u²du

Now we substitute, u = Kcos[t]
du = -Ksin[t]dt
And:
I/2 - K² * Sin^-1[u/K]
= -∫Ksin[t] * (-Ksin[t]dt)

CRAP THIS WASN'T EASIER [I'd still be doing the same thing ... oh well, I'll catch a break soon]
Thus we have:
(I/(K²) - 2Sin^-1[u/K])
=∫2sin²[t]dt

Now how the hell do you do a sin squared integral... Oh yeah...

(I/(K²) - 2Sin^-1[u/K])
=∫(1 - cos[2t])dt
=t - sin[2t]/2 + C₁

Now for some hectic rearrangement, gives us:

t = cos^-1[u/K] = cos^-1[Sqrt[(x-a)/(b-a)]]

sin[2t] = 2sin[t]cos[t]
= 2Sqrt[1 - cos²[t]]cos[t]
= 2Sqrt[1 - u²]u
= 2sqrt[1 - x - a]Sqrt[x - a]

So we've got:

(I/(K²) - 2Sin^-1[u/K])
= cos^-1[Sqrt[(x-a)/(b-a)]] - sqrt[1 - x - a]Sqrt[x - a] + C₁

Ok, this is literally the worst solution ever.

I/K²
= 2Sin^-1[Sqrt[(x-a)/(b-a)]] + cos^-1[Sqrt[(x-a)/(b-a)]] - sqrt[1 - x - a]Sqrt[x - a] + C₁

Thus:
I = (b - a) * [2Sin^-1[Sqrt[(x-a)/(b-a)]] + cos^-1[Sqrt[(x-a)/(b-a)]] - sqrt[1 - x - a]Sqrt[x - a] + C₁]

Yeah, that doesn't even look right. Someone find the error, I'm not liking the sqrt[1 - x - a] term, [it doesn't fit in]

 

[icycloud]

Keypad I haven't read over your whole solution yet, but there IS a better substitution . Shall I let other people attempt the question before divulging it?

 

[KeypadSDM]

Keypad I haven't read over your whole solution yet, but there IS a better substitution . Shall I let other people attempt the question before divulging it?

God yes. That just turned into question 8 2003. Whoops, dumb substitution there, now how do those integrals match?

Meh, fudge factor.

 

[icycloud]

Additional hint for those stuck: (don't look unless you've attempted the question first!!!)

Spoiler

Use a trigonometric substitution.

 

[KeypadSDM]

Additional hint for those stuck: (don't look unless you've attempted the question first!!!)

Spoiler

Use a trigonometric substitution.

What like

Spoiler

tan?

 

[icycloud]

What like

Spoiler

tan?

Nar. This would probably give it away.

Spoiler

One involving cos, sin, a and b.

 

[KeypadSDM]

Ok, nope. I have to say 2 years off really kills your 4 unit mad skillz.

 

[icycloud]

Ok, nope. I have to say 2 years off really kills your 4 unit mad skillz.

Hehe shall I PM you the substitution? =D

 

[KeypadSDM]

Hehe shall I PM you the substitution? =D

No way. If you do I will kill you. I'm not satisfied with an answer that's not mine.

 

[icycloud]

No way. If you do I will kill you. I'm not satisfied with an answer that's not mine.

Haha alright! Have fun with the question. The substitution is quite elegant.