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Grade: 12 

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

Answers : (1)

Arun
24742 Points 
							
∫ (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 
3 years ago
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