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Question
Evaluate: `int_0^(pi/2) (sin^2x)/(1 + cos x)^2 "d"x`
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Solution
Let I = `int_0^(pi/2) (sin^2x)/(1 + cos x)^2 "d"x`
Put `tan (x/2)` = t
∴ x = 2tan−1t
∴ dx = `(2"dt")/(1 + "t"^2)`, sin x `(2"t")/(1 + "t"^2)` and x = `(1 - "t"^2)/(1 + "t"^2)`
When x = 0, t = 0 and when x = `pi/2`, t = 1
∴ I = `int_0^1 ((2"t")/(1 + "t"^2))^2/(1 + (1 - "t"^2)/(1 + "t"^2))^2 * (2"dt")/(1 + "t"^2)`
= `int_0^1 ((4"t"^2)/(1 + "t"^2)^2)/(4/(1 + "t"^2)^2) * (2"dt")/(1 + "t"^2)`
= `2int_0^1 "t"^2/(1 + "t"^2) "dt"`
= `2int_0^1((1 + "t"^2 - 1)/(1 + "t"^2)) "dt"`
= `2int_0^1(1 + 1/(1 + "t"^2)) "dt"`
= `2["t" - tan^-1"t"]_0^1`
= 2[(1 – tan−11) − (0 − tan−10)]
= `2(1 - pi/4)`
= `(4 - pi)/2`
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