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∫9+x9-x dx - Mathematics and Statistics

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प्रश्न

`int sqrt((9 + x)/(9 - x))  "d"x`

योग
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उत्तर

Let I = `int  sqrt((9 + x)/(9 - x))  "d"x`

= `int sqrt((9 + x)/(9 - x) xx (9 + x)/(9 + x))   "d"x`

= `int (9 + x)/(sqrt((9)^2 -x^2))  "d"x`

= `int [9/sqrt((9)^2 - x^2) + x/sqrt((9)^2 - x^2)]  "d"x`

= `9 int 1/sqrt((9)^2 - x^2)  "d"x +int x/sqrt((9)^2 - x^2)  "d"x`

= `9 sin^(-1) (x/9) + "I"_1`

In I1, put (9)2 − x2 = t

∴ – 2x dx = dt

∴ x dx = `-1/2  "dt"`

∴ I1 = `-1/2  int  "dt"/sqrt("t")`

∴ = `-1/2 * (("t"^(1/2))/(1/2)) + "c"`

= `- sqrt(9^2 - x^2) + "c"`

∴ I = `9 sin^(-1)(x/9) - sqrt(81 - x^2) + "c"`

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अध्याय 2.3: Indefinite Integration - Short Answers II

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