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

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