4u Mathematics Marathon V 1.0 (2 Viewers)

[Riviet]

Alright, I'll post up the next one then:

Find ∫tan⁴x dx

 

[.ben]

New questiOn:

1+sinθ+icosθ
-------------- = sinθ + icosθ
1+sinθ-icosθ

hence deduce that:

[1+sin(∏/5)+cos(∏/5)]^5+[1+sin(∏/5)-cos(∏/5)]^5=0

edit:sorry riviet already posted a question

 

[Mountain.Dew]

Alright, I'll post up the next one then:

Find ∫tan⁴x dx

okay, here goes (this is going to be a LONG one, cant find any other more efficient way...)

let I = ∫tan⁴x dx

now, i use tan²x = sec²x - 1
therefore, tan⁴x = sec⁴x - 2sec²x + 1...squaring both sides.

so, I = ∫[sec⁴x - 2sec²x + 1] dx
I = ∫sec⁴x dx - [2tanx - x] + e, some constant e
now, let I = J - [2tanx - x] + e, where J = ∫sec⁴x dx

now, J = ∫sec⁴x dx
J = ∫[sec²x][sec²x] dx
now, using integration by parts:

u = sec²x, u' = 2secx[secxtanx] = 2sec²xtanx
v = tanx, v' = sec²x

so, using J = uv - ∫v u' dx,

J = tanx*sec²x - ∫2sec²xtan²x dx

J = tanx*sec²x - 2∫ [sec²x * [sec²x - 1] dx, using the identity tan²x = sec²x - 1

J = tanx*sec²x - 2∫sec⁴x dx + 2∫sec²x dx...expanding and spliting the integral into two integrals.

J = tanx*sec²x - 2J + 2 tanx + c

therefore, 3J = tanx*sec²x + 2 tanx + c,

J = 1/3 [ tanx*sec²x + 2 tanx ] + d, where c,d are constants.

THEREFORE, I = J - [2tanx - x] + e,

I = 1/3 [ tanx*sec²x + 2 tanx ] - 2tanx + x + f, some constant f.

yes, that was a mouthful...i will be humbled if you find a better method.

 

[Mountain.Dew]

ah, indeed, a BOS member - Stan.. - has found a better method to do the integral, which is remarkably more simplistic and is more preferred...

considering that tan⁴x = [tan²x][tan²x],

we only substitute one term --> tan⁴x = [tan²x][sec²x - 1]

so we get --> tan⁴x = tan²x*sec²x - tan²x

so therefore, ∫tan⁴x = ∫tan²x*sec²x dx - ∫tan²x dx.

then, by substitution of u = tanx, or by merely observing, we obtain more preferred solution:

∫tan⁴x = tan³x/3 - tanx + x + C

[mind you there is a possibility of more than one primitive of a certain function]

again, thank you to Stan.. for this! this post i owe to you. i am humbled.

 

[Mountain.Dew]

now, to continue this marathon...

find ∫ [(x-4)^1/2 / (5-2x)^1/2] dx

hint:

Spoiler

rationalise the numerator. *wink* *wink*

 

[icycloud]

i will be humbled if you find a better method.

Umm...

∫tan⁴x dx
= ∫(sec²x-1)tan²x dx
= ∫sec²xtan²x dx - ∫tan²x dx
= ∫tan²x d(tan x) - ∫sec²x-1 dx
= tan³x / 3 - tan x + x + C
#

 

[Mountain.Dew]

Umm...

∫tan⁴x dx
= ∫(sec²x-1)tan²x dx
= ∫sec²xtan²x dx - ∫tan²x dx
= ∫tan²x d(tan x) - ∫sec²x-1 dx
= tan³x / 3 - tan x + x + C
#

again, i am humbled...partying after the HSC certainly has its downsides.

thank you for ur post, icy cloud.

 

[Riviet]

That's the true power of direct integration. Nice work guys.

 

[Mountain.Dew]

New questiOn:

1+sinθ+icosθ
-------------- = sinθ + icosθ
1+sinθ-icosθ

hence deduce that:

[1+sin(∏/5)+cos(∏/5)]^5+[1+sin(∏/5)-cos(∏/5)]^5=0

edit:sorry riviet already posted a question

aaah might as well do it:

1+sinθ+icosθ
-------------- = sinθ + icosθ
1+sinθ-icosθ

now, consider RHS:

sinθ + icosθ = -i²sinθ + icosθ, since -i² = 1
then, factorising i out --> sinθ + icosθ = i[-isinθ + cosθ]
therefore, [sinθ + icosθ]⁵ = [i[cosθ-isinθ]]⁵
[sinθ + icosθ]⁵ = i *[cos5θ-isin5θ]....by De Movires theorem (dont know how to spell correctly, but anyway...)

so,
[1+sinθ+icosθ]⁵
------------------ = [sinθ + icosθ]⁵ = i *[cos5θ-isin5θ]
[1+sinθ-icosθ]⁵

THEN, sub θ = ∏/5

therefore,

[1+sin∏/5+icos∏/5]⁵
------------------------ = i *[cos5∏/5-isin5∏/5] = i[-1] = -i
[1+sin∏/5-icos∏/5]⁵

so, [1+sin∏/5+icos∏/5]⁵ = -i[1+sin∏/5-icos∏/5]⁵

then, [1+sin∏/5+icos∏/5]⁵ + i [1+sin∏/5-icos∏/5]⁵ = 0

here, i am stuck...

 

[icycloud]

now, to continue this marathon...

find ∫ [(x-4)^1/2 / (5-2x)^1/2] dx

I did this using trig. substitution. I'll leave the "rationalising the numerator" method for someone else .

Spoiler

I = ∫ √(x-4)/√(5-2x) dx
= ∫ √(x-4)/√(2(5/2 - x)) dx
= 1/√2 ∫ √(x-4) / √(5/2 - x) dx

Let x = 4cos²u + 5/2 sin²u
dx = -8cos(u)sin(u) + 5sin(u)cos(u) du
= -3sin(u)cos(u) du

Now,

I = -3/√2 ∫ √ (4cos²u + 5/2 sin²u - 4) / √ (5/2 - 4cos²u - 5/2sin²u) * sin(u)cos(u) du
= -3/√2∫√sin²u / √cos²u * sin(u)cos(u) du
= -3/√2∫sin²u du
= -3/(2√2)∫1-cos2u du
= -3/(2√2)[u - 1/2 sin2u] + C
= -3/(2√2)[u - sin(u)cos(u)] + C

Now, x = 4cos²u + 5/2 sin²u
= 4-4sin²u + 5/2sin²u
= -3/2 sin²u + 4

Thus, √(8-2x) / √3 = sin(u)
Also, √(2x-5) / √3 = cos(u)

Thus, I = -3/(2√2)[sin^-1(√(8-2x) / √3) - √(8-2x)(2x-5) / 3]

 

[Mountain.Dew]

WOW thats a method ive never seen before...interesting...

i was thinking of another method, but that seems fine, even if it does seem messy.

 

[icycloud]

WOW thats a method ive never seen before...interesting...

i was thinking of another method, but that seems fine, even if it does seem messy.

I would like to see your way . I thought my method was nice since it reduced down to the integral of Sin²u!

 

[Mountain.Dew]

I would like to see your way . I thought my method was nice since it reduced down to the integral of Sin²u!

behold, icycloud! (hehe, not really...i found out this is also messy, but can be neat if u got nice numbers, which i am sure the HSC will give u)

Spoiler

I = ∫ [(x-4)1/2 / (5-2x)1/2] dx
I = ∫(x-4) / ([(x-4)(5-2x)]1/2) dx ...times both sides by x-4
I = ∫(x-4) / ([5x-12-2x²]1/2) dx

then, complete the square with the denominator quadratic, split up the integral into x/ (quad) - 4/(quad) and use trigonometric substitution in the 1st integral, and the 2nd integral involves an inverse trig.

 

[icycloud]

behold, icycloud! (hehe, not really...i found out this is also messy, but can be neat if u got nice numbers, which i am sure the HSC will give u)

Haha yeh I did it that way too, but it's way messier! (it involves many more lines of working, even if the idea is much simpler than the other method) =D By the way, the bottom is meant to be (x-4)(5-2x) = - 2x^2 + 13x - 20

Anyway, you didn't give nice numbers!

p.s. And a challenge for everyone else to try and find another alternative method for this integral!

 

[Mountain.Dew]

mmmmm i guess it only works out right with nice numbers...oh well.

icy cloud, ur time to post up the next question.

 

[icycloud]

Alright, here goes:

∫Cos√x * Sin√x dx

Nothing too hard.

 

[Riviet]

Alright, here goes:

∫Cos√x * Sin√x dx

Nothing too hard.

Spoiler

∫Cos√x * Sin√x dx = 1/2.∫2sinx^1/2cosx^1/2 dx

= 1/2.∫(sin2x^1/2 + sin0) dx

= [cos2x^1/2]/2x^1/2 + C

Next Question:

Find ∫sin⁵x dx

 

[icycloud]

Hmm I don't think that's quite right. Can you expand on the working?

 

[Riviet]

Spoiler

I used the identity:

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

 

[Mountain.Dew]

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.

heres another method:

Spoiler

I = ∫Cos√x * Sin√x dx

let k = √x,. dk/dx = 1/2√x, so dx = dk 2√x = dk 2k

therefore, we get --> I =2∫k * cos k sin k dk = ∫k * sin2k dk

using integration by parts:

u = k, u' = 1
v =-1/2cos2k, v' = sin2k

I = -k + ∫1/2cos2k dk
I = 1/4sin2k - k + c
I = 1/4 sin2√x - √x + c