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प्रश्न
Evaluate the following:
`int_0^(pi/2) (tan x)/(1 + "m"^2 tan^2x) "d"x`
उत्तर
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)`.
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