Required integral =  sinx/(2sin2xcos2x) = _{[ ]}^{[ ]}  sinx/(4sinxcosxcos2x) =

_{[ ]}^{[ ]} dx/(4cosxcos2x) =

_{[ ]}^{[ ]} dx/(4cosx (1-2sin²x) ) = _{[ ]}^{[ ]} cosx/(4cos²x (1-2sin²x)) (Multiplying and dividing by cosx)

 

 =

_{[ ]}^{[ ]}  cosx/((1-sinx)(1+sinx)(1-2sin²x)) . Now take sinx = t => cosx = dt.

 

thus, the integral reduces to

_{[ ]}^{[ ]} dt/((1+t)(1-t)(1-2t²)) . On reducing this expression using partial fractions, I got the three constants A, B, C as -1/2, -1/2, 2. (Please notify if the values are wrong)

 

Thus, the integral reduces to :

 

_{[ ]}^{[ ]} -1/2 (dt/(1+t)) + [-1/2(dt/1-t)] + 2[dt/(1-2t²)].

 

The first two integrals are straightforward, and will be equal to log(1+t) and

log(1/(1-t)) respectively. For the third integral, there are many ways to solve it. You can either apply the direct formula for

_{[ ]}^{[ ]} dx/(a²-x²) = 1/2a (log(a+x)/(a-x)) or you can do it by completing the squares. Its your take!!

 

Cheers.