Question Number 100088 by mathmax by abdo last updated on 24/Jun/20

calculate ∫ ((cosx)/(cos(3x)))dx

Commented byDwaipayan Shikari last updated on 24/Jun/20

−(1/(2(√3)))log((((√3)sinθ−cosθ)/((√3)sinθ+cosθ)))+Constant

Answered by 1549442205 last updated on 25/Jun/20

cos 3x =4cos^3 x−3cos x ,so F=∫(dx/(4cos^2 x−3))=∫(dx/(2(1+cos 2x)−3))  =∫(dx/(2cos 2x −1))=(1/2)∫(dx/(cos 2x−(1/2)))  =_(t=tanx )  (1/2)∫(dt/((1+t^2 )(((1−t^2 )/(1+t^2 )))−(1/2))))=(1/2)∫(dt/(1−t^2 −((1+t^2 )/2)))  =∫(dt/(1−3t^2 ))=∫(dt/((1−(√3)t)(1+(√3)t)))=(1/2)∫((1/(1−(√3)t)))+(1/(1+(√3)t)))dt  =(1/2)∫(dt/(1+(√3)t))−(1/2)∫(dt/((√3)t−1))=(1/(2(√3)))∫((d(1+(√3)t))/(1+(√3)t))−(1/(2(√3)))∫((d((√3)t−1))/((√3)t−1))  =(1/(2(√3)))ln∣(((√3)t+1)/((√3)t−1))∣=(1/(2(√3)))ln∣(((√(3 ))arctan x+1)/((√3)arctan x−1))∣+C