# Hit me with a difficult P3 or P1 questionWatch

This discussion is closed.
14 years ago
#61
hmm try this

(tan x)^4 using substitution u=tan x
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#62
(Original post by powerball)
hmm try this

(tan x)^4 using substitution u=tan x
let u = tanx
du/dx = sec^2x
dx = du/sec^2x

{u^4. 1/sec^2x = u^5/5.sec^2x + c

(tanx)^5/5.sec^2x + c

Is that right?
0
14 years ago
#63
This one requires a bit of lateral thinking

∫ 1/(sin x cos x) dx
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14 years ago
#64
(Original post by BlueAngel)
let u = tanx
du/dx = sec^2x
dx = du/sec^2x

{u^4. 1/sec^2x = u^5/5.sec^2x + c

(tanx)^5/5.sec^2x + c

Is that right?
you have to integrate wrt. u, so you can't have that sec^2 x in there
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14 years ago
#65
answer is 1/3 tan^3 x -tan x + x if that gives you a clue to the method....
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14 years ago
#66
(Original post by powerball)
answer is 1/3 tan^3 x -tan x + x if that gives you a clue to the method....
this is a bit of a tough cookie
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14 years ago
#67
(Original post by powerball)
hmm try this

(tan x)^4 using substitution u=tan x
(tan^2 x)*(tan^2 x)

(tan^2 x)(sec^2 x - 1)

(tan^2 x)*(sec^2 x) - (tan^2 x)

then I did them separately

∫ (tan^2 x)*(sec^2 x)
u = tan x
du = sec^2 x

∫ u^2 du
= tan^3 x

tan^3 x - ( ∫ tan^2 x )

( ∫ tan^2 x )
= ∫ sec^2 x - 1
= tan x - x

tan^3 x - ( tan x - x )

tan^3 x - tan x + x
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14 years ago
#68
(Original post by kimoni)
(tan^2 x)*(tan^2 x)

(tan^2 x)(sec^2 x - 1)

(tan^2 x)*(sec^2 x) - (tan^2 x)

then I did them separately

∫ (tan^2 x)*(sec^2 x)
u = tan x
du = sec^2 x

∫ u^2 du
= tan^3 x

tan^3 x - ( ∫ tan^2 x )

( ∫ tan^2 x )
= ∫ sec^2 x - 1
= tan x - x

tan^3 x - ( tan x - x )

tan^3 x - tan x + x
Well done, ive been doing it for 5 mins but i got 1/5tan^5 x - 1/3tan^3 x + c again and again...need mroe practice me thinks!
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14 years ago
#69
(Original post by kimoni)
This one requires a bit of lateral thinking

∫ 1/(sin x cos x) dx
multipy cos x in the numerator and the denominator
and u 'll get INT cotx sec^2x
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#70
(Original post by kimoni)
This one requires a bit of lateral thinking

∫ 1/(sin x cos x) dx

let u = sinx
du/dx = cosx
dx = du/cosx

{1/ucosx.1/cosx .du = 1/u.cos^2x .du = 1/u.(1 - sin^2x)
{1/u - u^3 .du = {1/u - 1/u^3 .du = LNU - u^-2/2 + c

Is that right?
0
14 years ago
#71
(Original post by kimoni)
This one requires a bit of lateral thinking

∫ 1/(sin x cos x) dx
I = ∫ 1/(sinx.cosx) dx

u = sinx
du = cos x dx
dx = du/cosx
dx = du/(1-uÂ˛)

I = ∫ 1/u.(1-uÂ˛) du

1/u.(1-uÂ˛) = 1/u.(1+u)(1-u) = 1/u + Â˝/(1+u) - Â˝/(1-u)

I = ∫ 1/u du + Â˝ ∫ 1/(1+u) du - Â˝ ∫ 1/(1-u) du
I = lnu + Â˝ln(1+u) + Â˝ln(1-u) + lnA
I = lnAu(1-uÂ˛)^Â˝
===========
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14 years ago
#72
(Original post by Fermat)
dx = du/cosx
dx = du/(1-uÂ˛)
I think it should be

dx = du/sqrt(1-uÂ˛)
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14 years ago
#73
(Original post by mikesgt2)
I think it should be

dx = du/sqrt(1-uÂ˛)
Aaaaarg! So it is! No wonder it fell out
0
14 years ago
#74
Wait as minute. it still comes out!
u = sinx
du = cos x dx
dx = du/cosx

I = ∫ 1/(sinx.cosx) dx
I = ∫ 1/(sinx.cosx) du/cosx
I = ∫ 1/(sinx.cosÂ˛x) du
I = ∫ 1/u.(1-uÂ˛) du

just a typo
0
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