the integral 1/(a^2cos^2(x)+b^2sin^(x))^2 for the interval [0,pi/2] is

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Amlan Kumar · Answer

Divide the numerator and denominator by cos to the powe 4 x and than use an unique substitution atanx=btan t. On simplification your answer becomes pi(a sq+b sq)/a cube*b cube i hope you liked my answer

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the integral 1/(a^2cos^2(x)+b^2sin^(x))^2 for the interval [0,pi/2] is

the integral 1/(a^2cos^2(x)+b^2sin^(x))^2 for the interval [0,pi/2] is

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the integral 1/(a^2cos^2(x)+b^2sin^(x))^2 for the interval [0,pi/2] is

the integral 1/(a^2cos^2(x)+b^2sin^(x))^2 for the interval [0,pi/2] is

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