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Question
Evaluate the following : `int (1)/(cos2x + 3sin^2x).dx`
Solution
Let I = `int (1)/(cos2x + 3sin^2x).dx`
= `int (1)/(1 - 2sin^2x + 3sin^2x).dx`
= `int(1)/(1 + sin^2x).dx`
Dividing both numerator and denominator by cos2x, we get
I = `int(sec^2x dx)/(sec^2x + tan^2x)`
= `int (sec^2x dx)/(1 + tan^2x + tan^2x)`
= `int (sec^2x dx)/(2tan^2x + 1)`
Put tan x = t
∴ sec2x dx = dt
∴ I = `int (1)/(2t^2 + 1)dt`
= `(1)/(2) int (1)/(t^2 + (1/sqrt(2))^2)dt`
= `(1)/(2) xx (1)/((1/sqrt(2)))tan^-1 (t/(1/sqrt(2))) + c`
= `(1)/sqrt(2)tan^-1 (sqrt(2)tan x) + c`.
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