4u Mathematics Marathon V 1.0 (3 Viewers)

[icycloud]

Spoiler

I used the identity:

2sinAsinB=Sin(A+B) + Sin(A-B), does it work for A=B?

Sure, but how do you go from the 2nd to 3rd lines.

 

[icycloud]

heres another method:

Mm that's the method I had in mind, except you made a small mistake in the integration by parts.

 

[Riviet]

2sinAsinB=Sin(A+B) + Sin(A-B) <-- that is true, but it only works with both sins, we have a sin and a cos product, so i suppose u can use the other identity.

The three trig formula for Products to Sums or Differences are:

2sinAsinB=cos(A-B)-cos(A+B)
2cosAcosB=cos(A-B)+cos(A+B)
2sinAcosB=sin(A+B)+sin(A-B)

It was probably just your lost memory from that partying after hsc.

 

[Riviet]

Argh, I forgot how to integrate sin[f(x)]

I'll try it again tomorrow.

 

[icycloud]

Next Question:

Find ∫sin⁵ x dx

Spoiler

∫sin⁵x dx
= ∫(1-cos²x)² sin(x) dx
= ∫(1-2cos²x+cos⁴x) d(-cos(x))
= ∫2cos²x - 1 - cos⁴x d(cos(x))
= 2/3 cos³x - cos(x) - cos⁵x/5 + C
#

Next Question:
Find ∫dx/(1+sin²x)

 

[lfc_reds2003]

this man loves his maths..its great to see...

and he will dominate it for shizzy..GO TEAM

 

[Stan..]

Alright, here goes:

∫Cos√x * Sin√x dx

Nothing too hard.

1/2 ∫ sin 2x^1/2 dx u = 2x^1/2
-1/2 ∫u sinu du
-1/2 {u * -cosu - ∫-cosu du}
ucosu/2 - 1/2 sinu + C
x^1/2 cos 2x^1/2 - 1/2 sin 2x^1/2 + C

I made an error somewhere in there, too late to go through my working though.

 

[Stan..]

Next Question:
Find ∫dx/(1+sin²x)

Arctan (sinx) + C

 

[SeDaTeD]

Actually, that won't work since you'd also get a cosx once you differentiate that, by the chain rule.

 

[KeypadSDM]

Next Question:
Find ∫dx/(1+sin²x)

Kinda dodgy method (I honestly can't see a way with any simple substitutions, so I had to do some manipulative algebra to get some trig terms I wanted to see):

Spoiler

K = 1/(1+sin²x)
=1/(cos²x+2sin²x)
=sec²x/(1 + 2tan²x)

So that's the dodgy step. It's all downhill from there with a u = Sqrt[2]tan[x] substitution, or you can do it directly by parts:

K = (d/dx(tan[x]))/(1 + 2tan²[x])
= (1/Sqrt[2])(d/dx(Sqrt[2]tan[x]))/(1 + (Sqrt[2]tan[x])²)

Thus:
I = ∫Kdx = (1/Sqrt[2])∫(d/dx(Sqrt[2]tan[x]))/(1 + (Sqrt[2]tan[x])²)dx

Now we've got something of the form, f'(x)/(1 + [f(x)]²)
Which integrates to give tan^-1[f(x)]

Thus:
I = (1/Sqrt[2]) * tan^-1[Sqrt[2]tan[x]] + C
= tan^-1[Sqrt[2]tan[x]]/Sqrt[2] + C

A tad dodgy, but yeah, whatever.

Note that the original manipulation was quite lucky, I wouldn't have seen the substitution otherwise.

 

[icycloud]

Kinda dodgy method (I honestly can't see a way with any simple substitutions, so I had to do some manipulative algebra to get some trig terms I wanted to see):

Nice method. Here is another way for those interested:

Spoiler

1/(1+Sin[x]^2) = 1/(1 + (1/2 - Cos[2x]/2))
= 2/(3 - Cos[2x])

Then, use t = Tan[x], dt = t^2 + 1 dx
And proceed as with any t-substitution question, you should end up with the arctan integral dt/(2t^2+1), and come up with the same answer as Keypad's.

 

[KeypadSDM]

Oh yeah, the t method. I forgot about that little nugget. It makes everything so easy.

 

[icycloud]

Keypad, next question please . Preferably integration.

 

[KeypadSDM]

Keypad, next question please . Preferably integration.

I don't know any good ones, I just like doing them.

Fine, I'll just go and look up some past papers. Are you HAPPY NOW?

 

[Riviet]

You love integration don't you Aaron?

 

[Yip]

Here is a nice integration question: (sorry for the crap presentation, dunno how to input definite integrals into here)

Find ∫from -pi/4 to pi/4[[sin^2(x)]/[1+e^(-x)]dx]

Spoiler

Hint: Prove that ∫from -pi/4 to 0 [[sin^2(x)]/[1+e^(-x)]dx]=∫from 0 to pi/4 [[sin^2(x)]/[1+e^x]dx] and use this fact to evaluate the definite integral

 

[KeypadSDM]

Find ∫from -pi/4 to pi/4[[sin^2(x)]/[1+e^(-x)]dx]

Didn't use the spoiler, but here's my method:

Spoiler

set x = -y, and you'll find that:

I = ∫from -pi/4 to pi/4[[sin^2(x)]/[1+e^(-x)]dx]
=∫from -pi/4 to pi/4[[sin^2(y)]/[1+e^(y)]dx]

Thus, just looking at what's being integrated shows integrating the following two expressions is equivalent:

A(x) = sin²[x]/(1 + e^-x)
B(x) = sin²[x]/(1 + e^x)
= e^-xsin²[x]/(e^-x + 1)

That last one is by multiplying top and bottom by: e^-x

And so:

e^-xA(x) = B(x)

Therefore we have:
2I = ∫from -pi/4 to pi/4[A(x) + B(x)]dx
= ∫from -pi/4 to pi/4[(1 + e^-x)A(x)]dx
= ∫from -pi/4 to pi/4[sin²[x]]dx
= 2∫from 0 to pi/4[sin²[x]]dx

[the last line occurs as the function being integrated is even]

So we have:

I = ∫from 0 to pi/4[sin²[x]]dx
=0.5∫from 0 to pi/4[1 - cos[2x]]dx
=0.5[x - sin[2x]/2]from 0 to pi/4
=0.5(pi/4 - sin[pi/2]/2 - 0 + 0/2)
=0.5(- 1/2 + pi/4)
=pi/8 - 1/4

I reckon that's right, I haven't checked though. [Using one of the approximations]

And no, I can't be bothered thinking up another question. I'm no good at that.

 

[Yip]

Thats correct keypad ^^
It would be more efficient do it like this imo:

Spoiler

Prove that ∫from -pi/4 to 0 [[sin^2(x)]/[1+e^(-x)]dx]=∫from 0 to pi/4 [[sin^2(x)]/[1+e^x]dx] using the substitution x=-t
Then u split the integral into [-pi/4,0] and [0,pi/4], and u will get

I=∫from 0 to pi/4 [[[sin^2(x)]/[1+e^(-x)]]+[[sin^2(x)]/[1+e^x]]dx
=∫from 0 to pi/4[[(1+e^x)/(1+e^x)]sin^2(x)]dx
=pi/8-1/4

 

[SeDaTeD]

Went up some messy direction but reading the hint made it a bit too easy.

 

[icycloud]

Hey, just saw the question. Didn't look at the spoilers, here's my method:

Spoiler

Let I = ∫(-pi/4 --> pi/4) Sin[x]^2 / (1+E^x) dx
= ∫(-pi/4 --> pi/4) Sin[x]^2 E^x / (1+E^x) dx
= ∫(-pi/4 --> pi/4) F(x) dx

Now, J = ∫(-pi/4 --> pi/4) F(-x) dx
= ∫(-pi/4 --> pi/4) Sin[x]^2 / (1+E^x) dx

Now, let u = -x, du=-dx, we have:
J = ∫(-pi/4 --> pi/4) F(u) du
= I
= ∫(-pi/4 --> pi/4) Sin[x]^2 E^x / (1+E^x) dx
= ∫(-pi/4 --> pi/4) Sin[x]^2 / (1+E^x) dx

Thus, I + J = 2I = ∫(-pi/4 --> pi/4) Sin[x]^2 E^x / (1+E^x) dx + ∫(-pi/4 --> pi/4) Sin[x]^2 / (1+E^x) dx
= ∫(-pi/4 --> pi/4) Sin[x]^2 dx

Thus, I = 1/2∫(-pi/4 --> pi/4) Sin[x]^2 dx
= ∫(0 --> pi/4) Sin[x]^2 dx (symmetry)
= 1/2[x - 1/2Sin[2x]] (0 --> pi/4)
= pi/8 - 1/4
#

Edit: OK I read the other two guys' posts. Seems like my method is pretty much the same as Keypad's. Yip's way is more efficient mmm...

You love integration don't you Aaron?

Haha yeh, one of my favourite topics in 4U (maybe because it's so easy relative to the other topics hehe).