Let I=∫cot2x−tan2xcos8x+1dx Now, Dr=cot2x−tan2x=sin2xcos2x−cos2xsin2x =sin2xcos2xcos22x−sin22x=sin4x2cos4x ∴I=∫sin4x2cos4x2cos24xdx=∫2cos4x2cos24x⋅sin4xdx=21∫sin8xdx=−218cos8x+k=−161⋅cos8x+k Now ,−161⋅cos8x+k=Acos8x+k⇒A=−161
If the integral ∫cot2x−tan2xcos8x+1dx=Acos8x+k where k is an arbitrary constant, then A is equal to:
Held on 25 Apr 2013 · Verified 6 Jul 2026.
−161
161
81
−81
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