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प्रश्न
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
उत्तर
Put x – α = t2.
Then β – x = β – (t2 +α)
= β – t2 – α
= – t2 – α + β
And dx = 2tdt.
Now I = `int (2"t dt")/sqrt("t"^2(beta - alpha - "t"^2))`
= `int (2"dt")/sqrt((beta - alpha - "t"^2))`
= `2 "dt"/sqrt("k"^2 - "t"^2)`, where k2 = β – α
= `2sin^-1 "t"/"k" + "C"`
= `2sin^-1 sqrt((x - alpha)/(beta - alpha)) + "C"`
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