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Question
Evaluate the following:
`int_0^2 ("d"x)/("e"^x + "e"^-x)`
Solution
Let I = `int_0^2 ("d"x)/("e"^x + "e"^-x)`
= `int_0^1 ("d"x)/("e"^x + 1/"e"^x)`
= `int_0^1 ("d"x)/(("e"^(2x) + 1)/"e"^x)`
= `int_0^1 ("e"^x "d"x)/("e"^(2x) + 1)`
Put ex = t
⇒ ex dx = dt
Changing the limit, we have
When x = 0
∴ t = e0 = 1
When x = 1
∴ I = `int_1^"e" "dt"/("t"^2 + 1)`
= `[tan^-1 "t"]_1^"e"`
= `[tan^-1 "e" - tan^-1 (1)]`
= `tan^-1 "e" - pi/4`
Hence, I = `tan^-1 "e" - pi/4`.
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