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Answer»

∫ 1/(a²cos²x + b²sin²x) dx from 0 to π/2

= ∫ sec²x/(a² + b²tan²x) dx [dividing up and down by cos²x]

put btanx = u and sec²x dx = du/b. the integral becomes

(1/b) ∫ du/(a² + u²) from 0 to ∞

= (1/b) [(1/a) tanֿ¹(u/a)] from 0 to ∞

= (1/b) [(1/a) π/2]

= π/2ab.


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