Question

समाकलन  `int_(1/3)^1  (x - x^3)^(1/3)/x^4  dx` का मान है:

पर्याय

• 6

• 0

• 3

• 4

Answer 1

6

स्पष्टीकरण:

`x = cos theta` रखने पर,

`dx = cos theta  d theta`

`therefore int (x - x^3)^(1/3)/x^4  dx`

`= int ((sin theta - sin^3 theta)^(1/3))/(sin^4 theta)  cos theta  . d theta`

`= int (sin^(1/3) theta (1 - sin^2  theta)^(1/3))/(sin^4 theta)  cos theta  . d theta`

`= int (sin^(1/3) theta cos^(2/3) theta . cos theta)/(sin^2 theta sin^2 theta)`

`= int (cos^(5/3) theta)/(sin^(5/3) theta)  cosec^2 theta  d theta`

`= int cot^(5/3)  theta cosec^2  theta  d theta`

पुनः `cot theta = t` प्रतिस्थापित करने पर,

`⇒ -cosec^2 theta  d theta = dt`

`int (x - x^3)^(1/3)/x^4 = - int t^(5/3)  dt = (-3)/8  t^(8/3)`

`= (-3)/8  (cot theta)^(8/5)`

` = (-3)/8 ((cos theta)/(sin theta))^(8/3)`

`= (-3)/8 ((sqrt(1 - sin^2 theta))/sin theta)^(8/3)`

`= (-3)/8 [(sqrt(1 - x^2))/x]^(8/3)    ...[because sin theta = x]`

`therefore int_(1/3)^1 (x - x^3)^(1/3)/x^4  "dx" = (-3)/8 [((sqrt(1 - x^2))/x)^(8/3)]_(1/3)^1`

`=(-3)/8 [0 - ((sqrt(1 - 1/9))/(1/8))^(8/3)]`

`= 3/8 [((sqrt8)/3)/(1/3)]^(8/3) = 3/8 . (8^(1/2))^(8/3)`

`= 3/8 . 8^(8/6) = 3/8 * 2^(3 xx 8/6)`

`= 3/8 xx 2^4`

`= 3/8 xx 16 = 6`