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Question
Evaluate the following:
`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
Solution
Let 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))` .....[Making perfect square]
= `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`
Hence, I = `pi`.
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