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Question
Find `int "dx"/(2sin^2x + 5cos^2x)`
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Solution
Dividing numerator and denominator by cos2x, we have
I = `int (sec^2x "d"x)/(2tan^2x + 5)`
Put tanx = t
So that sec2x dx = dt.
Then I = `int "dt"/(2"t"^2 + 5) = 1/2 int "dt"/("t"^2 + (sqrt(5/2))^2`
= `1/2 sqrt(2)/sqrt(5) tan^-1 ((sqrt(2)"t")/sqrt(5)) + "C"`
= `1/sqrt(10) tan^-1 ((sqrt(2)tanx)/sqrt(5)) + "C"`.
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