Evaluate dx∫dx(x-α)(β-x),β>α

प्रश्न

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

बेरीज

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

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 5 Q 4 Q 6

पाठ 7: Integrals - Solved Examples [पृष्ठ १४८]

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पाठ 7 Integrals
Solved Examples | Q 5 | पृष्ठ १४८

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Evaluate dx∫dx(x-α)(β-x),β>α

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Evaluate dx∫dx(x-α)(β-x),β>α

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