Integral baholash usuli
Haqida maqolalar turkumining bir qismi |
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Yilda matematika, trigonometrik almashtirish bo'ladi almashtirish ning trigonometrik funktsiyalar boshqa iboralar uchun. Yilda hisob-kitob, trigonometrik almashtirish - bu integrallarni baholash texnikasi. Bundan tashqari, trigonometrik identifikatorlar aniq soddalashtirish integrallar o'z ichiga olgan radikal iboralar.[1][2] Almashtirish orqali boshqa integratsiya usullari singari, aniq integralni baholashda, integratsiya chegaralarini qo'llashdan oldin antiderivativni to'liq chiqarib olish osonroq bo'lishi mumkin.
I holat: o'z ichiga olgan integrallar ![{ displaystyle a ^ {2} -x ^ {2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/264648736234ef93254ab6ba708c519bbe1b5490)
Ruxsat bering
va foydalaning shaxsiyat
.
I holatiga misollar
I holat uchun geometrik qurilish
1-misol
Integral
![{ displaystyle int { frac {dx} { sqrt {a ^ {2} -x ^ {2}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e29e1b20bdd4c5cd53d6f81fd2b2e1042643d49)
biz foydalanishimiz mumkin
![{ displaystyle x = a sin theta, quad dx = a cos theta , d theta, quad theta = arcsin { frac {x} {a}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bad3b2a8cf8191aa4778676789581652d1bcedd3)
Keyin,
![{ displaystyle { begin {aligned} int { frac {dx} { sqrt {a ^ {2} -x ^ {2}}}} & = int { frac {a cos theta , d theta} { sqrt {a ^ {2} -a ^ {2} sin ^ {2} theta}}} [6pt] & = int { frac {a cos theta , d theta} { sqrt {a ^ {2} (1- sin ^ {2} theta)}}} [6pt] & = int { frac {a cos theta , d theta} { sqrt {a ^ {2} cos ^ {2} theta}}} [6pt] & = int d theta [6pt] & = theta + C [6pt] & = arcsin { frac {x} {a}} + C. end {aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb0f45f461035d567bc90912abb383b4f184bc87)
Yuqoridagi qadam shuni talab qiladi
va
. Biz tanlashimiz mumkin
ning asosiy ildizi bo'lish
va cheklovni joriy eting
teskari sinus funktsiyasidan foydalangan holda.
Aniq integral uchun integratsiya chegaralari qanday o'zgarishini aniqlash kerak. Masalan, kabi
dan ketadi
ga
, keyin
dan ketadi
ga
, shuning uchun
dan ketadi
ga
. Keyin,
![{ displaystyle int _ {0} ^ {a / 2} { frac {dx} { sqrt {a ^ {2} -x ^ {2}}}} = int _ {0} ^ { pi / 6} d theta = { frac { pi} {6}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2127bc2fad2ea93e68380a93642f51a93fcc91bd)
Chegaralarni tanlashda biroz ehtiyot bo'lish kerak. Chunki yuqoridagi integratsiya shuni talab qiladi
,
faqat borish mumkin
ga
. Ushbu cheklovni e'tiborsiz qoldirib, kimdir tanlagan bo'lishi mumkin
ketmoq
ga
, bu haqiqiy qiymatning salbiy tomoniga olib kelishi mumkin edi.
Shu bilan bir qatorda, chegara shartlarini qo'llashdan oldin noaniq integrallarni to'liq baholang. Bunday holda, antividiv vosita beradi
oldingi kabi.
2-misol
Integral
![{ displaystyle int { sqrt {a ^ {2} -x ^ {2}}} , dx,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de22d16c42bbe542699ff29f856928cf9a94a2c2)
ruxsat berish bilan baholanishi mumkin ![{ displaystyle x = a sin theta, , dx = a cos theta , d theta, , theta = arcsin { frac {x} {a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83db4f76bc98bd0ebf1955efda76aed226a39e2d)
qayerda
Shuning uchun; ... uchun; ... natijasida
va
kamon diapazoni bo'yicha, shunday qilib
va
.
Keyin,
![{ displaystyle { begin {aligned} int { sqrt {a ^ {2} -x ^ {2}}} , dx & = int { sqrt {a ^ {2} -a ^ {2} sin ^ {2} theta}} , (a cos theta) , d theta [6pt] & = int { sqrt {a ^ {2} (1- sin ^ {2} theta)}} , (a cos theta) , d theta [6pt] & = int { sqrt {a ^ {2} ( cos ^ {2} theta)}} , (a cos theta) , d theta [6pt] & = int (a cos theta) (a cos theta) , d theta [6pt] & = a ^ {2} int cos ^ {2} theta , d theta [6pt] & = a ^ {2} int left ({ frac {1+ cos 2 theta} {2} } o'ng) , d theta [6pt] & = { frac {a ^ {2}} {2}} chap ( theta + { frac {1} {2}} sin 2 theta right) + C [6pt] & = { frac {a ^ {2}} {2}} ( theta + sin theta cos theta) + C [6pt] & = { frac {a ^ {2}} {2}} left ( arcsin { frac {x} {a}} + { frac {x} {a}} { sqrt {1 - { frac {x ^ {2}} {a ^ {2}}}}} o'ng) + C [6pt] & = { frac {a ^ {2}} {2}} arcsin { frac {x} { a}} + { frac {x} {2}} { sqrt {a ^ {2} -x ^ {2}}} + C. end {aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6dc8b7727d973d3575d22f781010591f86e20436)
Aniq integral uchun, almashtirish amalga oshirilgandan so'ng chegaralar o'zgaradi va tenglama yordamida aniqlanadi
, oralig'idagi qiymatlar bilan
. Shu bilan bir qatorda, chegara atamalarini to'g'ridan-to'g'ri antidivivatsiya formulasiga qo'llang.
Masalan, aniq integral
![{ displaystyle int _ {- 1} ^ {1} { sqrt {4-x ^ {2}}} , dx,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b4600cfb267579e77528a004d8b67d61329b44d)
almashtirish bilan baholanishi mumkin
, yordamida belgilangan chegaralar bilan
.
Beri
va
,
![{ displaystyle { begin {aligned} int _ {- 1} ^ {1} { sqrt {4-x ^ {2}}} , dx & = int _ {- pi / 6} ^ { pi / 6} { sqrt {4-4 sin ^ {2} theta}} , (2 cos theta) , d theta [6pt] & = int _ {- pi / 6} ^ { pi / 6} { sqrt {4 (1- sin ^ {2} theta)}} , (2 cos theta) , d theta [6pt] & = int _ {- pi / 6} ^ { pi / 6} { sqrt {4 ( cos ^ {2} theta)}} , (2 cos theta) , d theta [ 6pt] & = int _ {- pi / 6} ^ { pi / 6} (2 cos theta) (2 cos theta) , d theta [6pt] & = 4 int _ {- pi / 6} ^ { pi / 6} cos ^ {2} theta , d theta [6pt] & = 4 int _ {- pi / 6} ^ { pi / 6} chap ({ frac {1+ cos 2 theta} {2}} o'ng) , d theta [6pt] & = 2 chap [ theta + { frac {1} {2}} sin 2 theta right] _ {- pi / 6} ^ { pi / 6} = [2 theta + sin 2 theta] { Biggl |} _ {- pi / 6} ^ { pi / 6} = chap ({ frac { pi} {3}} + sin { frac { pi} {3}} o'ng) - chap (- { frac { pi} {3}} + sin left (- { frac { pi} {3}} right) right) = { frac {2 pi} {3}} + { sqrt {3 }}. [6pt] end {hizalangan}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3290b5d8dffff518a7a54af50b0bbcad1051b19)
Boshqa tomondan, chegara atamalarini antivivativ hosil uchun ilgari olingan formulaga to'g'ridan-to'g'ri qo'llash
![{ displaystyle { begin {aligned} int _ {- 1} ^ {1} { sqrt {4-x ^ {2}}} , dx & = left [{ frac {2 ^ {2}} {2}} arcsin { frac {x} {2}} + { frac {x} {2}} { sqrt {2 ^ {2} -x ^ {2}}} right] _ {- 1} ^ {1} [6pt] & = chap (2 arcsin { frac {1} {2}} + { frac {1} {2}} { sqrt {4-1}} o'ng) - chap (2 arcsin chap (- { frac {1} {2}} o'ng) + { frac {-1} {2}} { sqrt {4-1}} o'ng) [6pt] & = chap (2 cdot { frac { pi} {6}} + { frac { sqrt {3}} {2}} o'ng) - chap (2 cdot chap (- { frac { pi} {6}} o'ng) - { frac { sqrt {3}} {2}} o'ng) [6pt] & = { frac {2 pi} {3}} + { sqrt {3}} end {aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/331bd80b5e0c5a19ece342b80e800bd3d1bc2093)
oldingi kabi.
II holat: o'z ichiga olgan integrallar ![{ displaystyle a ^ {2} + x ^ {2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fcd584d22bc92262f66dcbd397307df97494f76)
Ruxsat bering
va identifikatordan foydalaning
.
II holatga misollar
II holat uchun geometrik qurilish
1-misol
Integral
![{ displaystyle int { frac {dx} {a ^ {2} + x ^ {2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c09e16acbfd7c4c1079c381944b55247f8feada)
biz yozishimiz mumkin
![{ displaystyle x = a tan theta, quad dx = a sec ^ {2} theta , d theta, quad theta = arctan { frac {x} {a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f4dcc45c93a4a99ec39af3da242c8e3738f281d)
shunday qilib integral bo'ladi
![{ displaystyle { begin {aligned} int { frac {dx} {a ^ {2} + x ^ {2}}} & = int { frac {a sec ^ {2} theta , d theta} {a ^ {2} + a ^ {2} tan ^ {2} theta}} [6pt] & = int { frac {a sec ^ {2} theta , d theta} {a ^ {2} (1+ tan ^ {2} theta)}} [6pt] & = int { frac {a sec ^ {2} theta , d theta} {a ^ {2} sec ^ {2} theta}} [6pt] & = int { frac {d theta} {a}} [6pt] & = { frac { theta} {a}} + C [6pt] & = { frac {1} {a}} arctan { frac {x} {a}} + C, end {hizalangan}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c65e486a1f8cafb8397f72820972c35efacd858)
taqdim etilgan
.
Aniq integral uchun, almashtirish amalga oshirilgandan so'ng chegaralar o'zgaradi va tenglama yordamida aniqlanadi
, oralig'idagi qiymatlar bilan
. Shu bilan bir qatorda, chegara atamalarini to'g'ridan-to'g'ri antidivivatsiya formulasiga qo'llang.
Masalan, aniq integral
![{ displaystyle int _ {0} ^ {1} { frac {4} {1 + x ^ {2}}} , dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b360d6deec97cc74441299a307ca56518ff84505)
almashtirish bilan baholanishi mumkin
, yordamida belgilangan chegaralar bilan
.
Beri
va
,
![{ displaystyle { begin {aligned} int _ {0} ^ {1} { frac {4 , dx} {1 + x ^ {2}}} & = 4 int _ {0} ^ {1 } { frac {dx} {1 + x ^ {2}}} [6pt] & = 4 int _ {0} ^ { pi / 4} { frac { sec ^ {2} theta , d theta} {1+ tan ^ {2} theta}} [6pt] & = 4 int _ {0} ^ { pi / 4} { frac { sec ^ {2} theta , d theta} { sec ^ {2} theta}} [6pt] & = 4 int _ {0} ^ { pi / 4} d theta [6pt] & = (4 theta) { Bigg |} _ {0} ^ { pi / 4} = 4 chap ({ frac { pi} {4}} - 0 right) = pi. End {hizalangan }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1fdc8a13ac2312f87a1c7b36cef5ca23eb89075)
Shu bilan birga, antidiviv hosilning formulasiga chegara atamalarini to'g'ridan-to'g'ri qo'llash
![{ displaystyle { begin {aligned} int _ {0} ^ {1} { frac {4} {1 + x ^ {2}}} , dx & = 4 int _ {0} ^ {1} { frac {dx} {1 + x ^ {2}}} & = 4 chap [{ frac {1} {1}} arctan { frac {x} {1}} right] _ {0} ^ {1} & = 4 ( arctan x) { Bigg |} _ {0} ^ {1} & = 4 ( arctan 1- arctan 0) & = 4 chapga ({ frac { pi} {4}} - 0 o'ng) = pi, end {hizalanmış}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d22c46fc3be1aac3570a02e6914168f9e0fa0501)
oldingi kabi.
2-misol
Integral
![{ displaystyle int { sqrt {a ^ {2} + x ^ {2}}} , {dx}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b803094c2df957dda10a019daa3f7b0b552c54cf)
ruxsat berish bilan baholanishi mumkin ![{ displaystyle x = a tan theta, , dx = a sec ^ {2} theta , d theta, , theta = arctan { frac {x} {a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/becc82ec3cefef1516128cf00fdce37056c6516f)
qayerda
Shuning uchun; ... uchun; ... natijasida
va
Arktangens diapazoni bo'yicha, shuning uchun
va
.
Keyin,
![{ displaystyle { begin {aligned} int { sqrt {a ^ {2} + x ^ {2}}} , dx & = int { sqrt {a ^ {2} + a ^ {2} tan ^ {2} theta}} , (a sec ^ {2} theta) , d theta [6pt] & = int { sqrt {a ^ {2} (1+ tan ^ {2} theta)}} , (a sec ^ {2} theta) , d theta [6pt] & = int { sqrt {a ^ {2} sec ^ {2 } theta}} , (a sec ^ {2} theta) , d theta [6pt] & = int (a sec theta) (a sec ^ {2} theta) , d theta [6pt] & = a ^ {2} int sec ^ {3} theta , d theta. [6pt] end {aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/108a5f1becea83b5cb41021d81544ff3e1bab889)
The sekant kubikning ajralmas qismi yordamida baholanishi mumkin qismlar bo'yicha integratsiya. Natijada,
![{ displaystyle { begin {aligned} int { sqrt {a ^ {2} + x ^ {2}}} , dx & = { frac {a ^ {2}} {2}} ( sec theta tan theta + ln | sec theta + tan theta |) + C [6pt] & = { frac {a ^ {2}} {2}} left ({ sqrt {) 1 + { frac {x ^ {2}} {a ^ {2}}}}} cdot { frac {x} {a}} + ln left | { sqrt {1 + { frac { x ^ {2}} {a ^ {2}}}}} + { frac {x} {a}} right | right) + C [6pt] & = { frac {1} {2 }} left (x { sqrt {a ^ {2} + x ^ {2}}} + a ^ {2} ln left | { frac {x + { sqrt {a ^ {2} + x ^ {2}}}} {a}} right | right) + C. End {hizalangan}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35b28bc818f9ffcffedfb2e767d2d578c4a3e038)
III holat: o'z ichiga olgan integrallar ![{ displaystyle x ^ {2} -a ^ {2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13876905041dab46ee026a3ed803c28236efc5d6)
Ruxsat bering
va identifikatordan foydalaning ![{ displaystyle sec ^ {2} theta -1 = tan ^ {2} theta.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb4b3be9755542c047e4ecf6806046c37b05b628)
III holatga misollar
III holat uchun geometrik qurilish
Kabi integrallar
![{ displaystyle int { frac {dx} {x ^ {2} -a ^ {2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b30f75798064140e18d96a88c285fd35808652c)
tomonidan ham baholanishi mumkin qisman fraksiyalar trigonometrik almashtirishlar o'rniga. Biroq, ajralmas
![{ displaystyle int { sqrt {x ^ {2} -a ^ {2}}} , dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0adc36a27b13417d665fc5e270f00f76aef98dd)
qila olmaydi. Bunday holda, tegishli almashtirish:
![{ displaystyle x = a sec theta, , dx = a sec theta tan theta , d theta, , theta = operatorname {arcsec} { frac {x} {a}} ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb7821f86646643d27738332a5beabc1da05f84b)
qayerda
Shuning uchun; ... uchun; ... natijasida
va
taxmin qilish orqali
, Shuning uchun; ... uchun; ... natijasida
va
.
Keyin,
![{ displaystyle { begin {aligned} int { sqrt {x ^ {2} -a ^ {2}}} , dx & = int { sqrt {a ^ {2} sec ^ {2} theta -a ^ {2}}} cdot a sec theta tan theta , d theta & = int { sqrt {a ^ {2} ( sec ^ {2} theta - 1)}} cdot a sec theta tan theta , d theta & = int { sqrt {a ^ {2} tan ^ {2} theta}} cdot a sec theta tan theta , d theta & = int a ^ {2} sec theta tan ^ {2} theta , d theta & = a ^ {2} int ( sec theta) ( sec ^ {2} theta -1) , d theta & = a ^ {2} int ( sec ^ {3} theta - sec theta) , d theta. end {hizalangan}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfa25abbc33d5141fff3eebfe40b132b19709f60)
Kimdir buni baholashi mumkin sekant funktsiyasining ajralmas qismi raqamni va maxrajni ko'paytirish orqali
va sekant kubikning ajralmas qismi qismlar bo'yicha.[3] Natijada,
![{ displaystyle { begin {aligned} int { sqrt {x ^ {2} -a ^ {2}}} , dx & = { frac {a ^ {2}} {2}} ( sec theta tan theta + ln | sec theta + tan theta |) -a ^ {2} ln | sec theta + tan theta | + C [6pt] & = { frac {a ^ {2}} {2}} ( sec theta tan theta - ln | sec theta + tan theta |) + C [6pt] & = { frac {a ^ {2}} {2}} chap ({ frac {x} {a}} cdot { sqrt {{ frac {x ^ {2}} {a ^ {2}}} - 1}} - ln chap | { frac {x} {a}} + { sqrt {{ frac {x ^ {2}} {a ^ {2}}} - 1}} o'ng | o'ng) + C [6pt] & = { frac {1} {2}} chap (x { sqrt {x ^ {2} -a ^ {2}}} - a ^ {2} ln chap | { frac {x + { sqrt {x ^ {2} -a ^ {2}}}} {a}} right | right) + C. end {hizalangan}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d551bea9f1a33df981d45ab8cf11a1443d6da85)
Qachon
, bu qachon sodir bo'ladi
arcececant oralig'ini hisobga olgan holda,
, ma'no
o'rniga bu holda.
Trigonometrik funktsiyalarni yo'q qiladigan almashtirishlar
Almashtirish yordamida trigonometrik funktsiyalarni olib tashlash mumkin.
Masalan; misol uchun,
![{ displaystyle { begin {aligned} int f ( sin (x), cos (x)) , dx & = int { frac {1} { pm { sqrt {1-u ^ {2 }}}}} f chap (u, pm { sqrt {1-u ^ {2}}} o'ng) , du && u = sin (x) [6pt] int f ( sin ( x), cos (x)) , dx & = int { frac {1} { mp { sqrt {1-u ^ {2}}}}} f chap ( pm { sqrt {1) -u ^ {2}}}, u o'ng) , du && u = cos (x) [6pt] int f ( sin (x), cos (x)) , dx & = int { frac {2} {1 + u ^ {2}}} f chap ({ frac {2u} {1 + u ^ {2}}}, { frac {1-u ^ {2}} {1 + u ^ {2}}} right) , du && u = tan left ({ tfrac {x} {2}} right) [6pt] end {aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a9a11e89e8ccd82a402c1c24e5c755bdd6400a0)
Oxirgi almashtirish "sifatida tanilgan Weierstrassning almashtirilishi, qaysi foydalanishni qiladi tangens yarim burchakli formulalar.
Masalan,
![{ displaystyle { begin {aligned} int { frac {4 cos x} {(1+ cos x) ^ {3}}} , dx & = int { frac {2} {1 + u ^ {2}}} { frac {4 chap ({ frac {1-u ^ {2}} {1 + u ^ {2}}} o'ng)} { chap (1 + { frac { 1-u ^ {2}} {1 + u ^ {2}}} o'ng) ^ {3}}} , du = int (1-u ^ {2}) (1 + u ^ {2} ) , du & = int (1-u ^ {4}) , du = u - { frac {u ^ {5}} {5}} + C = tan { frac {x} {2}} - { frac {1} {5}} tan ^ {5} { frac {x} {2}} + C. end {aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a49ac1614b4b912a8521f37a3ff4c0aa9af07f78)
Giperbolik almashtirish
Ning almashtirishlari giperbolik funktsiyalar integrallarni soddalashtirish uchun ham foydalanish mumkin.[4]
Integral
, almashtirishni amalga oshiring
, ![{ displaystyle dx = a cosh u , du.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d8b28fe7197f9f8ced2a86378f9e43fc14c1840)
Keyin, identifikatorlardan foydalanib
va ![{ displaystyle sinh ^ {- 1} {x} = ln (x + { sqrt {x ^ {2} +1}}),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03c128d3d4baf90e60faf3cd920dd3681c82dd4e)
![{ displaystyle { begin {aligned} int { frac {1} { sqrt {a ^ {2} + x ^ {2}}}} , dx & = int { frac {a cosh u} { sqrt {a ^ {2} + a ^ {2} sinh ^ {2} u}}} , du [6pt] & = int { frac {a cosh {u}} {a { sqrt {1+ sinh ^ {2} {u}}}}} , du [6pt] & = int { frac {a cosh {u}} {a cosh u}} , du [6pt] & = u + C [6pt] & = sinh ^ {- 1} { frac {x} {a}} + C [6pt] & = ln left ( { sqrt {{ frac {x ^ {2}} {a ^ {2}}} + 1}} + { frac {x} {a}} right) + C [6pt] & = ln chap ({ frac {{ sqrt {x ^ {2} + a ^ {2}}} + x} {a}} o'ng) + C end {hizalanmış}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4de72234865476739112fe15f4849d934ebb1622)
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