Question

Evaluate the following:

`int_0^(pi/2) (tan x)/(1 + "m"^2 tan^2x) "d"x`

Answer 1

Let I = `int_0^(pi/2) (tan x)/(1 + "m"^2 tan^2x) "d"x`

= `int_0^(pi/2) (sinx/cosx)/(1 + "m"^2 (sin^2x)/(cos^2x)) "d"x`

= `int_0^(pi/2) (sinx/cosx)/((cos^2x + "m"^2 sin^2x)/cos^2x) "d"x`

= `int_0^(pi/2) (sin x cos x)/(cos^2x + "m"^2 sin^2x) "d"x`

= `int_0^(pi/2) (sinx cosx)/(1 - sin^2x + "m"^2 sin^2x) "d"x`

= `int_0^(pi/2) (sinx cosx)/(1 - sin^2x (1 - "m"^2)) "d"x`

Put sin²x = t

2 sin x cos x dx = dt

sin x cos x dx = `"dt"//2`

Changing the limits we get,

When x = 0

∴ t = sin²0 = 0

When x = `pi/2`

∴ t = `sin^2  pi/2` = 1

∴ I = `1/2 int_0^1  "dt"/(1 - (1 - "m"^2)"t")`

I = `1/2 int_0^1 "dt"/(1 + ("m"^2 - 1)"t")`

= `1/2 [(log [1 + "m"^2 - 1)"t")/("m"^2 - 1)]_0^1`

= `1/(2("m"^2 - 1)) [log(1 + "m"^2 - 1) - log(1)]`

= `(log|"m"^2|)/(2("m"^2 - 1))`

Hence, I = `(log|"m"^2|)/(2("m"^2 - 1)) = (log|"m"|)/("m"^2 - 1)`.