Evaluate: `int1/(a^2sin^2x+b^2cos^2x) dx`

1.	Evaluate: `int1/(a^2sin^2x+b^2cos^2x) dx`
Answer» Dividing the numerator and the denominator of the given integrand by `cos^(2)` x , we get `int(dx)/((a^(2)sin^(2)x+b^(2)cos^(2)x))=int(sec^(2)x)/(a^(2)tan^(2)x+b^(2))dx` `=int(dt)/((a^(2)t^(2)+b^(2)))` [ putting tan x = t] `=(1)/(a^(2))int(dt)/([t^(2)+((b)/(a))^(2)])=(1)/(a^(2))*(1)/((b/(a)))"tan"^(-1)(t)/((b/(a)))+C` `(1)/(ab)tan^(-1)((at)/(b))+C=(1)/(ab)tan^(-1)((atanx)/(b))+C`.

Discussion

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Evaluate: `int1/(a^2sin^2x+b^2cos^2x) dx`