Question

Evaluate the following:

`int_0^2 ("d"x)/("e"^x + "e"^-x)`

Answer 1

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 e^x = t

⇒ e^x 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`.