I know that : $$\int_0^{2\pi} dx/(a^2 \cos^2x+b^2 \sin^2x)^2=2\pi/ab$$
now : $$\int_0^{\pi} dx/(a \cos^2x+b \sin^2x)^2=?$$
my try:I do not know where to start.:(
thank you very much !
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Try dividing and multiplying by $b \cos^2( x)$. – Max Payne Dec 07 '16 at 10:39
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4An excellent answer has already been posted here http://math.stackexchange.com/questions/455147/closed-form-of-int-02-pi-fracdxa2-cos2xb2-sin2xn – Dec 07 '16 at 10:40
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@bacon might you give me a hand and mark this question as duplicate? – tired Dec 07 '16 at 11:07
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@tired, sure how do I do so? – Dec 07 '16 at 11:19
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@bacon click 'close' (below the main body of the question) and choose the option 'duplicate' – tired Dec 07 '16 at 11:21
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1@tired - ok no poblem :-) – Dec 07 '16 at 11:22