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Question
Evaluate: `int_0^(pi/2) 1/(5 + 4cos x) "d"x`
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Solution
Let I = `int_0^(pi/2) 1/(5 + 4cos x) "d"x`
Put `tan (x/2)` = t
∴ x = 2tan−1t
∴ dx = `2/(1 + "t"^2)` dt and cos x = `(1 - "t"^2)/(1 + "t"^2)`
When x = 0, t = 0 and when x = `pi/2`, t = 1
∴ I = `int_0^1 1/(5 + 4((1 - "t"^2)/(1 + "t"^2))) xx 2/(1 + "t"^2) "dt"`
= `2int_0^1 1/(5 + 5"t" + 4 - 4"t"^2) "dt"`
= `2int_0^1 1/(9 + "t"^2) "dt"`
= `2int_0^1 1/("t"^2 + 3^2) "dt"`
= `2[1/3 tan^-1("t"/3)]_0^1`
= `2/3[tan^-1(1/3) - tan^-1(0)]`
= `2/3 tan^-1 (1/3)`
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