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प्रश्न
Evaluate: `int_0^pi 1/(3 + 2sinx + cosx) "d"x`
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उत्तर
Let I = `int_0^pi 1/(3 + 2sinx + cosx) "d"x`
Put `tan (x/2)` = t
∴ x = 2tan−1t
∴ dx = `(2"dt")/(1 + "t"^2)`, sin x = `(2"t")/(1 + "t"^2)` and cos x = `(1 - "t"^2)/(1 + "t"^2)`
When x = 0, t = 0 and when x = π, t = ∞
∴ I = `int_0^∞ 1/(3 + 2((2"t")/(1 + "t"^2)) + (1 - "t"^2)/(1 + "t"^2)) xx (2 "dt")/(1 + "t"^2)`
= `int_0^∞ (2 "dt")/(3 + 3"t"^2 + 4"t" + 1 - "t"^2)`
= `int_0^∞ (2 "dt")/(2"t"^2 + 4"t" + 4)`
= `int_0^∞ "dt"/("t"^2 + 2"t" + 2)`
= `int_0^∞ "dt"/("t"^2 + 2"t" + 1 + 1)`
= `int_0^∞ "dt"/(("t" + 1)^2 + 1^2)`
= `[tan^-1 ("t" + 1)]_0^∞`
= `tan^-1(1 + ∞) - tan^-1(1 + 0)`
= `tan^-1(∞) - tan^-1 (1)`
= `pi/2 - pi/4`
∴ I = = `pi/4`
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