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प्रश्न
Integrate the following functions w.r.t. x : `int (1)/(3 - 2cos 2x).dx`
उत्तर
Let I = `int (1)/(3 - 2cos 2x).dx`
Put tan x = t
∴ x = tan–1 t
∴ dx = `dt/(1 + t^2) and cos2x = (1 - t^2)/(1 + t^2)`
∴ I = `int (1)/(3 - 2((1 - t^2)/(1 + t^2))).dt/(1 + t^2)`
= `int (1 + t^2)/(3 + 3t^2 - 2 + 2t^2).dt/(1 + t^2)`
= `int (1)/(1 + 5t^2)dt`
= `(1)/(5) int (1)/((1 /sqrt(5))^2 + t^2)dt`
= `(1)/(5) xx (1)/((1/sqrt(5)))tan^-1(t/(1/sqrt(5))) + c`
= `(1)/sqrt(5)tan^-1(sqrt(5)tanx) + c`.
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