Integrate cos4x-1/cotx-tanx .kindly give me answer please help me

Question

Arun · Answer

∫ (cos4x -1/(cotx-tanx)) dx = ∫ (cos4x -1/(cosx/sinx - sinx/cosx)) dx =

= ∫ (cos4x -1/(cos^2x - sin^2x)/sinx.cosx)) dx = ∫ (cos4x -1/(cos2x/sin2x/2)) dx =

= ∫ (cos4x -sin2x/2cos2x) dx = ∫ (cos4x - tan2x/2) dx =

= -sin4x/4 + ln(cosx) + c

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Integrate cos4x-1/cotx-tanx .kindly give me answer please help me

Integrate cos4x-1/cotx-tanx .kindly give me answer please help me

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