Fanyeni hii intergration waungwana

[QUALIFIED]

S√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√sinx)-------eqn(ii)
Substituting ---eqn(i)&(ii) into --eqn(a)
SVdU=x√sinx - 1/2S(xcosx)dx/(√sinx)
For S(xcosx)dx/(√sinx)
By By-Part tena
SUdV=UV-SVdU
Hapa let
U=x
=>dU=dx
And
dV=(cosx)/(√sinx)
Integrating both sides
=>V=2√sinx (kama hujaelewa hapo basi wewe ni kilaza)



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

mkuu naona matunda ya pale kileleni chin ya usimamiz wa engeneer

 

[NingaR]

mkuu naona matunda ya pale kileleni chin ya usimamiz wa engeneer

aiseee engineer alikua nowma atii mpaka giza jamaa anag'ang'ania kufundisha kweli kila mtu na fani yake aisee. alaf kumbe jamaa alipigaga mechanical engineering, sasa ndo akawa ana apply kwetu.
but all in all here we are.

Mungu ibariki Tanzania
Mungu ibaliki Kilele……

 

[Kijakazi]

S√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√sinx)-------eqn(ii)
Substituting ---eqn(i)&(ii) into --eqn(a)
SVdU=x√sinx - 1/2S(xcosx)dx/(√sinx)
For S(xcosx)dx/(√sinx)
By By-Part tena
SUdV=UV-SVdU
Hapa let
U=x
=>dU=dx
And
dV=(cosx)/(√sinx)
Integrating both sides
=>V=2√sinx (kama hujaelewa hapo basi wewe ni kilaza)



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

Duh mkuu twende taratibu sasa kashfa za nini tena? si watu wanajifunza kulikuwa hamna umuhimu wa maneno yote haya!

 

[QUALIFIED]

aiseee engineer alikua nowma atii mpaka giza jamaa anag'ang'ania kufundisha kweli kila mtu na fani yake aisee. alaf kumbe jamaa alipigaga mechanical engineering, sasa ndo akawa ana apply kwetu.
but all in all here we are.

Mungu ibariki Tanzania
mungu ibaliki kilele××××

haha ni kwere mpaka jina likawa mathematics institute ni mwendo wa 6am-7pm

 

[NingaR]

Duh mkuu twende taratibu sasa kashfa za nini tena? si watu wanajifunza kulikuwa hamna umuhimu wa maneno yote haya!

Kijana nakumbuka ticha wangu alikua ana piga swali juujuu ukimuuliza mwalimu hapo sijakupata yaani anaacha kufundisha anakugeukia utakula mabango mpaka basii
ata kwambia huko O-Level ulipitaje? uli ibiaeee???
namaswali mengineyo.
Yuludi kwenye mada mpaka mtu ufikie swali hilo inamaana umekomaa integration ndo maana nikasema kama huja elewa mpaka simple method niliyo tumia basi utakua hujui integration (ie. KILAZA WA INTEGRATION)

Kwani Mkuu nimekuacha sehemu??

 

[NingaR]

haha ni kwere mpaka jina likawa mathematics institute ni mwendo wa 6am-7pm

Mathematics College bhana!!! Alikua ananiua pale muda wa msosi ukifika anakwambia madogo tupige kipindi msosi ntawanunulia kipindi kikiisha!!

 

[Kijakazi]

Kijana nakumbuka ticha wangu alikua ana piga swali juujuu ukimuuliza mwalimu hapo sijakupata yaani anaacha kufundisha anakugeukia utakula mabango mpaka basii
ata kwambia huko O-Level ulipitaje? uli ibiaeee???
namaswali mengineyo.
Yuludi kwenye mada mpaka mtu ufikie swali hilo inamaana umekomaa integration ndo maana nikasema kama huja elewa mpaka simple method niliyo tumia basi utakua hujui integration (ie. KILAZA WA INTEGRATION)

Kwani Mkuu nimekuacha sehemu??

Mimi kwangu hiyo ni integration rahisi sana, sema saa nyingine mtu unakuwa hutambui impact ya kitu unachokisema au kukiandika kwa mtu mwingine haswa pale mtu anapokuwa ameuliza swali, kumbuka in mathematics we all have our gaps, kwa maana hapo sio swala tu kuintegrate ni lazima pia ajue kutumia differentiation by chain rule ili kufikia jibu na kama alikuwa hafahamu anaweza akarudi na kuziba hayo mapengo inaezekana tutiane moyo tu wote tunejenga nyumba moja!

 

[JPM605]

NingaR Hizo letting zina utata. Let u=x nadhani sio sahihi because hiyo x iko ndani ya underoot ya sin x. Sijui umeitoaje nje labda ungelet u iwe kitu chote haf dv iwe dx. Hata hivyo sijui utafanya mara ngapi ili liishe. Nangeshukuru kama ungerudi na kutoa ufafanuzi wa ziada.

 

[Kijakazi]

NingaR Hizo letting zina utata. Let u=x nadhani sio sahihi because hiyo x iko ndani ya underoot ya sin x. Sijui umeitoaje nje labda ungelet u iwe kitu chote haf dv iwe dx. Hata hivyo sijui utafanya mara ngapi ili liishe. Nangeshukuru kama ungerudi na kutoa ufafanuzi wa ziada.

Hajatoa x nje alichokifanya nikwamba S (sin x) sqr dx ni sawa tu na kuandika S 1* (sinx) sqr dx sasa ukiIntegrate S1dx = x: ndipo x ilipotoka

 

[fortunho]

S√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√sinx)-------eqn(ii)
Substituting ---eqn(i)&(ii) into --eqn(a)
SVdU=x√sinx - 1/2S(xcosx)dx/(√sinx)
For S(xcosx)dx/(√sinx)
By By-Part tena
SUdV=UV-SVdU
Hapa let
U=x
=>dU=dx
And
dV=(cosx)/(√sinx)
Integrating both sides
=>V=2√sinx (kama hujaelewa hapo basi wewe ni kilaza)



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

vilaza bhana

 

[NingaR]

Mimi kwangu hiyo ni integration rahisi sana, sema saa nyingine mtu unakuwa hutambui impact ya kitu unachokisema au kukiandika kwa mtu mwingine haswa pale mtu anapokuwa ameuliza swali, kumbuka in mathematics we all have our gaps, kwa maana hapo sio swala tu kuintegrate ni lazima pia ajue kutumia differentiation by chain rule ili kufikia jibu na kama alikuwa hafahamu anaweza akarudi na kuziba hayo mapengo inaezekana tutiane moyo tu wote tunejenga nyumba moja!

nimekupata mkuu

 

[NingaR]

NingaR Hizo letting zina utata. Let u=x nadhani sio sahihi because hiyo x iko ndani ya underoot ya sin x. Sijui umeitoaje nje labda ungelet u iwe kitu chote haf dv iwe dx. Hata hivyo sijui utafanya mara ngapi ili liishe. Nangeshukuru kama ungerudi na kutoa ufafanuzi wa ziada.

Mkuu awali mime let dU=dx
nilipo integrate ndo ikaja U=x
baada ya kusolve jibu likaja na integration nyingine ndo nika let U=x
pia By-Part methot hua inatabia ya kutoa jibu reeefu sana linaweza toa hata zaid ya term 10 so hua tunachukua term 3 za mwanzo + K(constant)

 

[NingaR]

vilaza bhana

una maana gani??

 

[JPM605]

Kijakazi kwenye S(underoot sin x)dx sio sahihi kusema u=x ila ingepaswa kuwa dv=dx ambayo finnaly inaleta v=x utata unakuja kwenye u ukiilet iwe lote liiilobaki inaweza chukua steps nyingi kweli sijui utaifupisha vp!

 

[Kijakazi]

Kijakazi kwenye S(underoot sin x)dx sio sahihi kusema u=x ila ingepaswa kuwa dv=dx ambayo finnaly inaleta v=x utata unakuja kwenye u ukiilet iwe lote liiilobaki inaweza chukua steps nyingi kweli sijui utaifupisha vp!

Mimi sijui taizo liko wapi Integration by parts:

Su'v= uv-S u*v' dx

S1*(Sinx)sqr dx = x*(sinx)sqr - S x * d/dx (sin x)sqr dx
kusolve hiyo d/dx unatummia chain rule, halafu unaendelea na I.b.p kama ilivyoosheshwa tayari na mchangiaji mwingine au shida iko wapi?

 

[NingaR]

Kijakazi kwenye S(underoot sin x)dx sio sahihi kusema u=x ila ingepaswa kuwa dv=dx ambayo finnaly inaleta v=x utata unakuja kwenye u ukiilet iwe lote liiilobaki inaweza chukua steps nyingi kweli sijui utaifupisha vp!

Mkuu huja elewa nini nime let dV=dx kwa qn uliyo toa nilipo fika mbele ikatokea integral nyingine ndo nikalet dU=x(hii sio ya swali ulilo uliza bali ni la integral iliyo jitokeza)
mkuu unanihuzunika kuona umashunfwa kutambua concept ndogo kama hii

 

[Moyibi]

Mkuu kwani VP??

naona hujamaliza Kazi

 

[simplemind]

Substitute sinx=t hence dx=dt/cosx. Using the identity cosx^2+sinx^2=1 integral is equal to integral sqrt(t)/(sqrt(1-t^2) with respect to t. Rationalizing the denominator you end up with elliptic integral with solution -2E(pi-2x)/4|2 + contant where E is elliptic integral of second order!

 

[Erick1]

expand sinX kwa series expansion method , kisha tumia tylor theorem kuondoa square-root then intergret

 

[njiwa]

Nisaidieni 'intergral of the Squareroot of Sin X with respect to X' Ahsanteni.

usiumize kichwa kuna on-line calculators now days nimejaribu hapo

Wolfram Mathematica Online Integrator