Integration Questions (2 Viewers)

[richz]

hello guys, i need some help with the following questions. All these are from the cambridge book

1) diagnostic test 5 question 18

2) integral (tan x + cot x) dx

3) integral (square root of (4- x^2))

4) Integral (1/x sec^2(inx) dx)

thnx in advance

 

[shafqat]

2) remember the standard form for logs : f'x/fx. so the answer is logsinx - log cosx = logtanx

 

[FinalFantasy]

hello guys, i need some help with the following questions. All these are from the cambridge book

1) diagnostic test 5 question 18

2) integral (tan x + cot x) dx

3) integral (square root of (4- x^2))

4) Integral (1/x sec^2(inx) dx)

thnx in advance

2) integral (tan x + cot x) dx
=int. (sinx\cosx +cos x\sinx) dx
=-ln|cos x|+ln |sin x| +C

3) integral (square root of (4- x^2))
let x=2sin@
dx=2cos@ d@
int. sqrt(4-x²) dx=int. 2sqrt(cos²@) 2cos@ d@=int. 4cos²@ d@
=2int. (cos2@+1)d@=2 (1\2 sin2@+@) +C=sin [2sin^-1 (x\2)]+sin^-1 (x\2) +C

4) Integral (1/x sec^2(inx) dx)
=int. 1\x sec² (lnx) dx
let u=lnx
du\dx=1\x
=int. 1\x sec² (lnx) dx=int. sec² u du
=tan u+C=tan (lnx)+C

 

[KFunk]

hello guys, i need some help with the following questions. All these are from the cambridge book

1) diagnostic test 5 question 18

18. If I_n = [1-->e] ∫ (lnx)ⁿdx, show that I_n = e - I_n-1


I_n = ∫ (lnx)ⁿdx = ∫ 1.(lnx)ⁿdx so use integration by parts where:

∫u'v = uv - ∫uv'

u' = 1 -------> u = x

v = (lnx)ⁿ= yⁿ (where y = lnx)
dv/dx = dv/dy.dy/dx = n.y^n-1.1/x
hence v' = n/x.(lnx)^n-1

I_n= [x(lnx)ⁿ]{between 1 and e} - ∫ x.n/x.(lnx)^n-1
= e - n∫(lnx)^n-1 (The e is there because ln1 = 0, lne = 1 so the first bracket just becomes 'e')

I_n= e - nI_n-1

To find I₄ then just use the reccurance trick. (EDIT: may as well finish it)

I₄ = e - 4I₃
= e - 4(e - 3I₂)
=-3e + 12(e - 2I₁)
= 9e - 24I₁ [I₀ is just ∫1dx between 1 and e which equals (e -1)]
= 9e - 24(e -(e-1))
= 9e -24

 

[richz]

3) integral (square root of (4- x^2))
let x=2sin@
dx=2cos@ d@
int. sqrt(4-x²) dx=int. 2sqrt(cos²@) 2cos@ d@=int. 4cos²@ d@
=2int. (cos2@+1)d@=2 (1\2 sin2@+@) +C=sin [2sin^-1 (x\2)]+sin^-1 (x\2) +C

Thnx guys

PS final fantasy i think it shud be 2 (sin^-1 (x\2)) instead of sin^-1 (x\2)

 

[FinalFantasy]

oh yea.. at the end lol
oops!

 

[richz]

ok how about this one

integral (square root of (16 + x^2)

 

[richz]

sorry guys, to bug u again i also need help with the following

Cambridge further questions 5 q 46-49

thnx

 

[KFunk]

Haha interesting timing, I did all of those yesterday. I'll write up what I can be bothered to do, for the most part I'll try and just give you hints because it's best if you can try and do them yourself. Give me 20 mins or so.

 

[Slidey]

ok how about this one

integral (square root of (16 + x^2)

u=16+x^2
du=2xdx

1/2 * Integral sqrt(u) du

u^(3/2)/3 + C

(16+x^2)^3/2/3 + C

 

[richz]

ok thnx kfunk yeah i dont really want u to give me the answwers i just want sum hints.. thnx

 

[richz]

u=16+x^2
du=2xdx

1/2 * Integral sqrt(u) du

u^(3/2)/3 + C

(16+x^2)^3/2/3 + C

sorry slide i dont think that agrees with the answers..

 

[KFunk]

47. I_n = [0-->π/2] ∫ xⁿsinxdx

do the ∫u'v = uv -∫ uv' where u'=sinx and v=xⁿ

I_n = [-xⁿcosx]{between 0-->π/2} + n∫x^n-1cosx dx

= [0-->π/2] n∫x^n-1cosx dx

then use the same process of integration by parts on this integral and you should end up with:
I_n= n(π/2)^n-1 -n(n-1)I_n-2

If that still bothers you, give us a yell.

 

[KFunk]

49. Have a go at substitutions like (u=π/2 - x) or (u = x - π/2) and see how it works out.

Keep in mind that [-a --->a]∫f(x)dx = - [a --->-a] ∫f(x)dx and vice versa

 

[richz]

kool 47 was good thnx

 

[richz]

ok k funk that 47 was all good thnx.. oppps thnx for ya help for 49, accidently included it

 

[KFunk]

For 48. Integration by parts:

∫ u'v = uv - ∫ uv' where u' = cosx and v=cos^n-1x

Remember the identity sin²x = 1 - cos²x and don't hesitate to rearange I_n and I_n-2 terms

 

[richz]

for 48.

is the next line -(n-1) int. (sinxcos^(n-2)x) dx??

 

[shafqat]

yeah sliderule theres no x
substitute x = 4tanp

 

[richz]

ook thnx shaf