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Evaluate dx∫dx(x-α)(β-x),β>α - Mathematics

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प्रश्न

Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`

योग
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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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Definite Integrals
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अध्याय 7: Integrals - Solved Examples [पृष्ठ १४८]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 7 Integrals
Solved Examples | Q 5 | पृष्ठ १४८

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