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Question
Evaluate: `int_0^(pi/4) (sec^2x)/(3tan^2x + 4tan x + 1) "d"x`
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Solution
Let I = `int_0^(pi/4) (sec^2x)/(3tan^2x + 4tan x + 1) "d"x`
Put tan x = t
∴ sec2x dx = dt
When x = 0, t = 0 and when x = `pi/4`, t = 1
∴ I = `int_0^1 "dt"/(3"t"^2 + 4"t" + 1)`
= `1/3 int_0^1 "dt"/("t"^2 + (4"t")/3 + 1/3)`
= `1/3 int_0^1 "dt"/("t"^2 + 2((2"t")/3) + (2/3)^2 - (2/3)^2 + 1/3)`
= `1/3 int_0^1 "dt"/(("t" + 2/3)^2 + ((-4 + 3)/9))`
= `1/3 int_0^1 "dt"/(("t" + 2/3)^2 - (1/3)^2`
= `1/3[1/(2 xx 1/3) log|(("t" + 2/3) - 1/3)/(("t" + 2/3) + 1/3)|]_0^1`
= `1/2[log|(3"t" + 1)/(3"t" + 3)|]_0^1`
= `1/2[log(4/6) - log(1/3)]`
= `1/2 log(4/6 xx 3)`
∴ I = `1/2 log 2`
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