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\[\int\limits \cos ^2x \div \cos ^2x+4\sin ^2x dx\]
so 1+4sin^2(x)?
any one to help me
denominator has (cos^2x+4sin^2x)?
yes
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sin^2x=(1-cosx)/2 cos^2x=(1+cosx)/2 replace and solve
wait i do
is cos^2x=(cos2x+1)/2 not (1+cosx)/2
sry its i was wrong replace x by 2x and 2x by x in lhs and rhs respectively
we get \[\int\limits ( \cos2x+1)/5\cos2x-3\]
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so cn u help me how to finish t?
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