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Evaluate: ∫0π4 sec2x3tan2x+4tanx+1 dx - Mathematics and Statistics

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Question

Evaluate: `int_0^(pi/4)  (sec^2x)/(3tan^2x + 4tan x + 1)  "d"x`

Sum
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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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Methods of Evaluation and Properties of Definite Integral
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Chapter 2.4: Definite Integration - Long Answers III

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