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Question
Evaluate the following : `int (1)/(4 + 3cos^2x).dx`
Solution
Let I = `int (1)/(4 + 3cos^2x).dx`
Dividing both numerator and denominator by cos2x, we get
I = `int (sec^2x)/(4sec^2 x + 3).dx`
= `int (sec^2x)/(4(1 + tan^2x) + 3).dx`
= `int (sec^2x)/(4tan^2x + 7).dx`
Put tan x = t
∴ sec2x dx = dt
I = `int dt/(4t^2 + 7)`
= `int dt/((2t)^2 + (sqrt(7))^2`
= `(1)/sqrt(7)tan^-1 ((2t)/sqrt(7)).(1)/(2) + c`
= `(1)/(2sqrt(7))tan^-1 ((2tanx)/sqrt(7)) + c`.
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