integral  cosx/(1+sin2x) dx 1+sin2x = (sinx+cosx)2 = 1/2 (cosx+sinx+cosx-sinx)/(sinx+cosx) 2 dx split into 2 integrals first one 1/(sinx+cosx) dx put tanx/2 =t dx = 2dt /1+t2 sinx = 2t/1+t2 cosx = 1-t2/1+t2 use this second one (cosx-sinx)/sinx+cosx) 2 put sinx+cosx =t numerator is dt thus u can proceed to solve the sum
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