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Evaluate: ∫121 (ecos-1x)(sin-1x)1-x2 dx - Mathematics and Statistics

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Question

Evaluate: `int_(1/sqrt(2))^1  (("e"^(cos^-1x))(sin^-1x))/sqrt(1 - x^2)  "d"x`

Sum
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Solution

Let I = `int_(1/sqrt(2))^1  (("e"^(cos^-1x))(sin^-1x))/sqrt(1 - x^2)  "d"x`

= `int_(1/sqrt(2))^1 (("e"^(pi/2 - sin^-1x))(sin^-1x))/sqrt(1 - x^2)  "d"x`   ......`[∵ sin^-1x + cos^-1x = pi/2]`

Put sin−1x = t

∴ `1/sqrt(1 - x^2)  "d"x` = dt

When x = `1/sqrt(2)`, t = `pi/4` and when x = 1, t = `pi/2`

∴ I = `int_(pi/4)^(pi/2) ("e"^(pi/2 - "t"))"t"  "dt"`

= `["t" int"e"^(pi/2 - "t") "dt"]_(pi/4)^(pi/2) - int_(pi/4)^(pi/2)["d"/("dt") ("t") int"e"^(pi/2 - "t")  "dt"]  "dt"`

= `["t"* ("e"^(pi/2 - "t"))/(-1)]_(pi/4)^(pi/2) - int_(pi/4)^(pi/2) 1* ("e"^(pi/2 - "t"))/(-1)  "dt"`

= `-(pi/2 "e"^0 - pi/4 "e"^(pi/4)) + int_(pi/4)^(pi/2) "e"^(pi/2 - "t")  "dt"`

= `-(pi/2 - pi/4 "e"^(pi/4)) + [("e"^(pi/2 - "t"))/(-1)]_(pi/4)^(pi/2)`

= `- pi/2 + pi/4 "e"^(pi/4) - ("e"^0 - "e"^(pi/4))`

= `- pi/2 + pi/4 "e"^(pi/4) - 1 + "e"^(pi/4)`

∴ I = `"e"^(pi/4) (pi/4 + 1) - (pi/2 + 1)`

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Methods of Evaluation and Properties of Definite Integral
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Chapter 2.4: Definite Integration - Long Answers III

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