Tentang integral, butuh bantuan?

misalkan x = 3sec y

maka sec y = x/3

dan

dx = 3 sec y.tan y dy

x² - 9 = (3sec y)² - 9 = 9sec²y - 9 = 9(sec²y -1) = 9tan²y

maka soal berubah menjadi

∫ (x² - 9)^(3/2) dx

= ∫ (9 tan² y)^(3/2) 3.sec y.tan y dy

= ∫ 27.tan³ y . 3.sec y. tan y dy

= ∫ 81 tan² y. tan² y. sec y dy

= ∫ 81. (sec² y - 1)² . sec y dy

= ∫ 81 (sec^4 - 2 sec² y + 1) sec y dy

= ∫ 81 (sec^5 - 2 sec^3 y + secy) dy

lalu gunakan rumus reduksi, yang ribet dan susah diketikkan di sini.

Untuk lebih jelas, lihat saja situs ini:

http://www.wolframalpha.com/input/?i=integrate+%28...

ini saya copykan langkah2nya dari situs tersebut!

Possible intermediate steps:

integral (-9+x^2)^(3/2) dx

For the integrand, (x^2-9)^(3/2) substitute x = 3 sec(u) and dx = 3 tan(u) sec(u) du. Then (x^2-9)^(3/2) = (9 sec^2(u)-9)^(3/2) = 27 tan^3(u) and u = sec^(-1)(x/3):

= 81 integral tan^4(u) sec(u) du

For the integrand tan^4(u) sec(u), use the trigonometric identity tan^2(u) = sec^2(u)-1:

= 81 integral sec(u) (sec^2(u)-1)^2 du

Expanding the integrand sec(u) (sec^2(u)-1)^2 gives sec^5(u)-2 sec^3(u)+sec(u):

= 81 integral (sec^5(u)-2 sec^3(u)+sec(u)) du

Integrate the sum term by term and factor out constants:

= 81 integral sec^5(u) du-162 integral sec^3(u) du+81 integral sec(u) du

Use the reduction formula, integral sec^m(u) du = (sin(u) sec^(m-1)(u))/(m-1) + (m-2)/(m-1) integral sec^(-2+m)(u) du, where m = 3:

= 81 integral sec^5(u) du-81 tan(u) sec(u)

Use the reduction formula, integral sec^m(u) du = (sin(u) sec^(m-1)(u))/(m-1) + (m-2)/(m-1) integral sec^(-2+m)(u) du, where m = 5:

= 81/4 tan(u) sec^3(u)-81 tan(u) sec(u)+243/4 integral sec^3(u) du

Use the reduction formula, integral sec^m(u) du = (sin(u) sec^(m-1)(u))/(m-1) + (m-2)/(m-1) integral sec^(-2+m)(u) du, where m = 3:

= 81/4 tan(u) sec^3(u)-405/8 tan(u) sec(u)+243/8 integral sec(u) du

The integral of sec(u) is log(tan(u)+sec(u)):

= 81/4 tan(u) sec^3(u)-405/8 tan(u) sec(u)+243/8 log(tan(u)+sec(u))+constant

The integral of sec(u) is log(tan(u)+sec(u)):

= 81/4 tan(u) sec^3(u)-405/8 tan(u) sec(u)+81 integral sec(u) du-405/8 log(tan(u)+sec(u))

The integral of sec(u) is log(tan(u)+sec(u)):

= 81/4 tan(u) sec^3(u)-405/8 tan(u) sec(u)+243/8 log(tan(u)+sec(u))+constant

Substitute back for u = sec^(-1)(x/3):

= 1/4 sqrt(x^2-9) (x^2)^(3/2)-45/8 sqrt(x^2-9) sqrt(x^2)+243/8 log(1/3 (sqrt(1-9/x^2)+1) x)+constant

Factor the answer a different way:

= 1/8 (sqrt(x^2) sqrt(x^2-9) (2 x^2-45)+243 log(1/3 (sqrt(1-9/x^2)+1) x))+constant

Which is equivalent for restricted x values to:

= 1/8 (x sqrt(x^2-9) (2 x^2-45)+243 log(2 (sqrt(x^2-9)+x)))+constant

∫ (x*2-9)*2 (3/2) dx

Misalkan.

u = x2-9

du = 2x dx

dx = 1/2x du

maka.

∫ (x*2-9)*2 (3/2) dx

∫ u*2 (3/2) 1/2x du

1/3 u*3 (3/2 x) 1/2x + c

1/3 u*3 (3/4) + c

3/12 u*3 + c

¼ u*3 + c

(¼ x2–9)*3 + c

*2=pggkat 2

*3=pgkt 3

Maaf gua pake hape.,, jadi susah ngtiknya.,,

Moga benar uaaa.,, amin..,

J = ∫ (x² - 9)^(3/2) dx

misalkan

x = 3 cosh u

dx = 3 sinh u du

J = ∫ (9 cosh² u - 9)^(3/2)(3 sinh u du)

J = ∫ 81 sinh⁴ u du

J = ∫ 81 (sinh² u)² du

J = ∫ 81 (½ cosh 2u - ½)² du

J = ∫ 81 (¼ cosh² 2u - ½ cosh 2u + ¼) du

J = ∫ 81 (¼ (½ cosh 4u + ½) - ½ cosh 2u + ¼) du

J = (-81/8) ∫(4 cosh (2u) - cosh (4u) - 3) du

J = (-81/8) (2 sinh (2u) - ¼ sinh (4u) - 3u) + C

Ubah lagi u menjadi bentuk x, sehingga J merupakan fungsi dari x.

bisa juga dalam bentuk fungsi yang lain, yang penting jika dimasukkan nilai batas atas dan batas bawah, akan menghasilkan nilai yang sama.