Question

integrate 1/ (a2cos2x + b2sin2​x) dx from 0 to pi/2

Answer 1

1/(a²cos2x +b²sin2x)dx


a²cos2x +b²sin2x = √(a⁴+b⁴){a²/√(a⁴+b⁴).cos2x+b²/√(a⁴+b⁴)sin2x}

=√(a⁴+b⁴)sin(2x + tan^-1(a²/b²) )

put this in integrant,

1/√(a⁴+b⁴) . dx/sin(2x +tan^-1(a²/b²))

let tan^-1(a²/b²) =∅

=1/√(a⁴+b⁴) .cosec(2x+∅) dx

=1/2√(a⁴+b⁴)ln|cosec(2x+∅) -cot(2x+∅)| +C
where C is constant ,