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प्रश्न
Evaluate : `int_0^pi (1)/(3 + 2sinx + cosx)*dx`
बेरीज
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उत्तर
Let I = `int_0^pi (1)/(3 + 2sinx + cosx)*dx`
Put `tan x/(2)` = t
∴ x = 2 tan–1 t
∴ dx = `(2dt)/(1 + t^2)`
and
sinx = `(2t)/(1 + t^2), cos x = (1 - t^2)/(1 + t^2)`
When x = 0, t = tan0 = 0
When x = `pi, t = tan pi/(2) = oo`
∴ I = `int_0^oo (1)/(3 + 2((2t)/(1 + t^2)) + ((1 - t^2)/(1 + t^2)))*(2dt)/(1 + t^2)`
= `int_0^oo (1)/(2t^2 + 4t + 4)*dt`
= `(2)/(2) int_0^oo (1)/(t^2 + 2t + 2)*dt`
= `int_0^oo (1)/((t^2 + 2t + 1) + 1)*dt`
= `int_0^oo (1)/((t^2 + 2t + 1 + 1)*dt`
= `int_0^oo (1)/((t + 1)^2 + (1)^2)*dt`
= `(1)/(1)[tan^-1 ((t + 1)/1)]_0^oo`
= `[tan^-1 (t + 1)]_0^oo`
= `tan^-1 oo - tan^1 1`
= `pi/(2) - pi/(4)`
= `pi/(4)`
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