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प्रश्न
Integrate the following functions w.r.t. x : `int (1)/(2sin 2x - 3)dx`
उत्तर
Let I = `int (1)/(2sin 2x - 3)dx`
Put tan x = t
∴ x = tan–1 t
∴ dx = `dt/(1 + t^2) and sin 2x = (2t)/(1 + t^2)`
∴ I = `int(1)/(2((2t)/(1 + t^2)) - 3).dt/(1 + t^2)`
= `int (1 + t^2)/(4t - 3 - 3t^2).dt/(1 + t^2)`
= `int (1)/(-3t^2 + 4t - 3)dt`
= `(1)/(3) int (1)/(t^2 - 4/3t + 1)dt`
= `-(1)/(3) int (1)/((t^2 - 4/3t + 4/9) - (4)/(9) + 1)dt`
= `-(1)/(3) int (1)/((t - 2/3)^2 + (sqrt(5)/3)^2)dt`
= `-(1)/(3) xx (1)/((sqrt(5)/3))tan^-1 ((t - 2/3)/(sqrt(5)/3)) + c`
= `-(1)/sqrt(5)tan^-1 ((3t - 2)/sqrt(5)) + c`
= `-(1)/sqrt(5)tan^-1((3tan x - 2)/(sqrt(5))) + c`.
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