# Fanyeni hii intergration waungwana

Discussion in 'Jukwaa la Elimu (Education Forum)' started by JPM605, Sep 10, 2012.

1. J

### JPM605Senior Member

#1
Sep 10, 2012
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Nisaidieni 'intergral of the Squareroot of Sin X with respect to X' Ahsanteni.

2. ### Don DavidJF-Expert Member

#2
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Pure 2......

3. ### King Kong IIIJF-Expert Member

#3
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Cosx/2

4. ### Mbwiga88JF-Expert Member

#4
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jaribu kutumia concept ya half angle

5. B

### Boniface92Member

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taylor theorem

6. K

### KitukoJF-Expert Member

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Duh ndonga yangu imeshajaa maisha ambayo hayana uhusiano kabisa na hiyo intergration, enzi za mwalimu hiyo nilikuwa nakushukia kwa kichwa tu, haaaa ,,, SO MUCH SIMPLE

7. ### AutorunJF-Expert Member

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=-cos x+c

8. ### ze duduzJF-Expert Member

#8
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Nilikuwa CBG na nilikuwa nakula BAM Ila uwezo wa kupiga Banda la Pure Maths nilikuwa nao Xo jibu mpango mzma tumia taylor serie na jibu ni =x2/4-x4/64+c
Note: x2 is x power 2 and x4 is x power 4

9. ### KijakaziJF-Expert Member

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Sep 10, 2012
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Nafikiri unaweza ukatumia Substitution method

S (sinx) X (SinX) dx (S naitumia kama integral sign) hapo nimeexpand tu sin x squared
halafu fanya substitution z= sinx
z'= cos
dz/dx=cosx
dx= 1/cos dz
Halafu unaweka kwenye sinx = z; S z (sinx)dx

Baada ya hapo substutute dx; S z(sinx)1/sinx dz
sinx na sinx zinajicancel unabakia na S z dz hapo sasa kazi imeisha unaintegrate tu z dz
S z dz=1/2 z(squared) +C halafu unafanya resubstitution yaani pale pa z = sin x
(sinx)squared/2 + C na hilo ndilo jibu lake (kumbuka nimetumia S kama integral sign!)

10. f

### fortunhoSenior Member

#10
Sep 11, 2012
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nimeamin hesabu sio mapishi.dogo fanya hivi,hauwezi ukapata exact answer na ndio maana wakilitoa huwaga wanaweka limits ilimradi utumie simpson au trapezoidal,akikuuliza aproxmat ans bila limit ndo unamkumbuka taylor.

11. ### NingaRJF-Expert Member

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kamawewe unajua hesabu tumia By-Part method then acha jibu katika 3 terms

12. ### IpycalypseJF-Expert Member

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Kumbe ulikua huamini.

13. f

### fortunhoSenior Member

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haya bhana

14. ### paul kiterejaJF-Expert Member

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Sep 11, 2012
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tatizo hilo!

15. J

### JPM605Senior Member

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Halina limits wala number of recquired ordinates hivyo naona trapezoidal na simpson zinafail..., haf tailor inaaply if f(x)=(x+a) kitu ambayo hii function haina.., kuhusu by parts nisaidieni ni ipi itakuwa U na dv?

16. ### NingaRJF-Expert Member

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Sep 11, 2012
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S&#8730;sinx.dx
Hapa hakuna ujanja zaidi ya kutumia By-Part method.
ie. SVdU=UV-SUdV----eqn(a)
let dU=dx
Integrating both side
=>U=x------eqn(i)
and V=(sinx)^½
differentiating both sides
=>dV=(cosx)/(2&#8730;sinx)-------eqn(ii)
Substituting ---eqn(i)&(ii) into --eqn(a)
SVdU=x&#8730;sinx - 1/2S(xcosx)dx/(&#8730;sinx)
For S(xcosx)dx/(&#8730;sinx)
By By-Part tena
SUdV=UV-SVdU
Hapa let
U=x
=>dU=dx
And
dV=(cosx)/(&#8730;sinx)
Integrating both sides
=>V=2&#8730;sinx (kama hujaelewa hapo basi wewe ni ******)

NIMECHOKA KUANDIKA
NATHANI MTAKUA MMEPATA MWANGA
(KAMA HUJUI HESABU KAA KUSHOTO)

17. ### ze duduzJF-Expert Member

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Sep 11, 2012
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teh teh kaka ulipata ngapi? Namba

18. ### NingaRJF-Expert Member

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Sep 11, 2012
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Mkuu kwani VP??

19. ### tindikalikaliJF-Expert Member

#19
Sep 11, 2012
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