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Putting x=alphacos^(2)theta+betasin^(2)t...

Putting `x=alphacos^(2)theta+betasin^(2)theta` show that, `int_(alpha)^(beta)sqrt((x-alpha)(beta-x))dx=((beta-alpha)^(2)pi)/(8)`

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The correct Answer is:
`((beta-alpha)^(2)pi)/(8)`
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