Evaluate 
∫cosec2x(cos2x-sin2x)2dx
Opening the square
On multiplying (1-sin2x)(1-sin2x) equation reduces to
=∫(cosec2x-2+sin2x-2cos2x+ sin2x)dx
=∫(cosec2x-2+2sin2x-2cos2x)dx
=∫(-2(cos2x-sin2x)+cosec2x-2)dx
=∫(-2cos2x+cosec2x-2)dx
On solving this we get our answer i.e
=-sin2x-cotx-2x+c
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