Question

निम्नलिखित का मान निकालिए-

`int_1^2 ("d"x)/sqrt((x -1) (2 -x))`

Answer 1

मान लीजिए I = `int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`

= `int_1^2 ("d"x)/sqrt(2x - x^2 - 2 + x)`

= `int_1^2 ("d"x)/sqrt(-x^2 + 3x - 2)`

= `int_1^2 ("d"x)/sqrt(-(x^2 - 3x + 2)`

= `int_1^2 ("d"x)/sqrt(-(x^2 - 3x + 9/4 - 9/4 + 2))`  .....[पूर्ण वर्ग बनाना]

= `int_1^2 ("d"x)/sqrt(-[(x - 3/2)^2 - 1/4])`

= `int_1^2 ("dx)/sqrt(1/4 - (x - 3/2)^2)`

= `int_1^2 ("d"x)/sqrt((1/2)^2 - (x - 3/2)^2)`

= `[sin^-1 ((x - 3/2)/(1/2))]_1^2`

= `[sin^-1 ((2x - 3)/1)]_1^2`

= `sin^-1 (4 - 3) - sin^-1 (2 - 3)`

= `sin^-1 (1) - sin^-1 (-1)`

= `sin^-1 (1) + sin^-1 (1)`

 = `2 sin^-1 (1)`

= `2 xx pi/2`

= `pi`

अत:, I = `pi`