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