integration of dx/(((cosx)^2 +4sinx-5)cosx)
asked by guest
on Jan 25, 2025 at 9:55 pm
You asked:
Evaluate the integral:
∫(cos(x)2+4sin(x)−5)⋅cos(x)1dx
MathBot Answer:
Antiderivative
∫(cos(x)2+4sin(x)−5)⋅cos(x)1dx=−18−18tan(2x)+18tan2(2x)8log(1+tan2(2x)−tan(2x))−18−18tan(2x)+18tan2(2x)2log(1+tan(2x))+18−18tan(2x)+18tan2(2x)3tan(2x)+18−18tan(2x)+18tan2(2x)18log(−1+tan(2x))−18−18tan(2x)+18tan2(2x)18log(−1+tan(2x))tan(2x)−18−18tan(2x)+18tan2(2x)8tan2(2x)log(1+tan2(2x)−tan(2x))−18−18tan(2x)+18tan2(2x)2tan2(2x)log(1+tan(2x))+18−18tan(2x)+18tan2(2x)2log(1+tan(2x))tan(2x)+18−18tan(2x)+18tan2(2x)8log(1+tan2(2x)−tan(2x))tan(2x)+18−18tan(2x)+18tan2(2x)18tan2(2x)log(−1+tan(2x))+C