Evaluate 
∫ sec2x (cos2x – sin2x )2 dx
Opening the square
On multiplying 
equation reduces to
=∫(cos2x-2sin2x+sec2x-2+cos2x)dx
=∫(2cos2x-2sin2x+sec2x-2)dx
=∫(2(cos2x-sin2x)+sec2x-2)dx
=∫(2cos2x+sec2x-2)dx
On solving this we get our answer i.e
=sin2x+tanx-2x+c
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