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Question
`int_0^1 1/(sqrt(3 + 2x - x^2))dx` = ______
Options
`pi/6`
`pi/3`
`pi/4`
`pi/2`
MCQ
Fill in the Blanks
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Solution
`int_0^1 1/(sqrt(3 + 2x - x^2))dx` = `underline(pi/6)`
Explanation:
`int_0^1 1/(sqrt(3 + 2x - x^2))dx`
= `int_0^1 1/(sqrt(4 - 1 + 2x - x^2))dx`
= `int_0^1 1/(sqrt((2)^2 - (x - 1)^2))dx = [sin^-1((x - 1)/2)]_0^1`
= `sin^-1(0) - sin^-1(-1/2) = 0 - (-pi/6)`
= `pi/6`
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Some Special Integrals
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