Math, asked by KINGARTHUR, 11 months ago

integral of 6x-5 root of 6-2x^2+xdx evaluate

Answers

Answered by amitnrw
8

Given : (6x - 5)√(6 - 2x²  + x)dx

To Find : Integrate

Solution:

(6x - 5)√(6 - 2x²  + x)

= (3/2)(4x  - 10/3)√(6 - 2x²  + x)

= (3/2)(4x -1 + 1 - 10/3)√(6 - 2x²  + x)

= (3/2)(4x -1   - 7/3)√(6 - 2x²  + x)

= (3/2)(4x -1)√(6 - 2x²  + x)   - (7/2)√(6 - 2x²  + x)

I = I1 - I2

I1 = ∫(3/2)(4x -1)√(6 - 2x²  + x) dx

(6 - 2x²  + x) = t

=> (-4x + 1)dx = dt

=> (4x - 1)dx = - dt

I1 = -∫(3/2) √ t dt

I1 = - (3/2) t^(3/2) / (3/2) + c1

I1 = - (6 - 2x²  + x)^(3/2) + c1

I2 = ∫√2 (7/2)√(3 -  x²  + x/2)dx

I2 = ∫√2 (7/2)√(3+ 1/16 - 1/16 -  x²  + x/2)dx

I2 =∫√2 (7/2)√(49/16 - ( (1/4)² + x²  -2.x(1/4))dx

I2 = (7/√2)∫√((7/4)² - ( x -1/4)²)dx

x -1/4 = y

dx = dy

I2 = (7/√2)∫√((7/4)² - ( y)²)dy

I2 = (7/√2) {  (y/2)√((7/4)² - ( y)²  + (7/4)²/2 sin⁻¹(y/(7/4)) + c2

substitute y = x -1/4

I2 = (7/√2) {  ((x -1/4)/2)√((7/4)² - ( x -1/4)²  + (7/4)²/2 sin⁻¹((x -1/4)/(7/4)) + c2

I2 = (7/√2) {  ((4x -1 )/8)√(3 -  x²  + x/2) + (49/32) sin⁻¹((4x -1)/7) + c2

I2 =  (7 (4x -1 )/16)√(6 -  2x²  + x ) + (343/32√2) sin⁻¹((4x -1)/7) + c2

I = I1 - I2

= - (6 - 2x²  + x)^(3/2) - (7 (4x -1 )/16)√(6 -  2x²  + x )  -  (343/32√2) sin⁻¹((4x -1)/7) +C

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