Evaluate : (i) `int(sinx)/((1-4cos^(2)x))dx`
(ii) `int("cosec"^(2)x)/((1-cot^(2)x))dx`

Explanation

(i) Putting `cosx=t" and "-sinx dx=dt`, we get
`int(sinx)/((1-4cos^(2)x))dx=-int(dt)/((1-4t^(2)))`
`=-(1)/(4)*int(dt)/(((1)/(4)-t^(2)))=(1)/(4)*int(dt)/({((1)/(2))^(2)-t^(2)})`
`=-(1)/(4)xx(1)/((2xx(1)/(2)))log|((1)/(2)+t)/((1)/(2)-t)|+C`
`=-(1)/(4)log|(1+2t)/(1-2t)|+C=-(1)/(4)log|(1+2cosx)/(1-cosx)|+C`.
(ii) Putting `cotx=t" and -cosec"^(2)xdx=dt` we get
`int("cosec"^(2)x)/((1-cot^(2)x))dx=-int(dt)/((1-t^(2)))=-(1)/((2xx1))log|(1+t)/(1-t)|+C`
`=-(1)/(2)log|(1+cotx)/(1-cotx)|+C`.