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

calculate ∫ ((cosx)/(cos(3x)))dx

$$\mathrm{calculate}\:\int\:\frac{\mathrm{cosx}}{\mathrm{cos}\left(\mathrm{3x}\right)}\mathrm{dx} \\ $$

Commented by Dwaipayan Shikari last updated on 24/Jun/20

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

$$−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}{log}\left(\frac{\sqrt{\mathrm{3}}{sin}\theta−{cos}\theta}{\sqrt{\mathrm{3}}{sin}\theta+{cos}\theta}\right)+{Constant} \\ $$

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

$$\mathrm{cos}\:\mathrm{3x}\:=\mathrm{4cos}^{\mathrm{3}} \mathrm{x}−\mathrm{3cos}\:\mathrm{x}\:,\mathrm{so}\:\mathrm{F}=\int\frac{\mathrm{dx}}{\mathrm{4cos}^{\mathrm{2}} \mathrm{x}−\mathrm{3}}=\int\frac{\mathrm{dx}}{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2x}\right)−\mathrm{3}} \\ $$$$=\int\frac{\mathrm{dx}}{\mathrm{2cos}\:\mathrm{2x}\:−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{dx}}{\mathrm{cos}\:\mathrm{2x}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\underset{\mathrm{t}=\mathrm{tanx}\:} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\left.\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}−\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{dt}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} −\frac{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$=\int\frac{\mathrm{dt}}{\mathrm{1}−\mathrm{3t}^{\mathrm{2}} }=\int\frac{\mathrm{dt}}{\left(\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{t}\right)\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{t}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\left.\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{t}\right)}+\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{t}}\right)\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{dt}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{t}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{dt}}{\sqrt{\mathrm{3}}\mathrm{t}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\int\frac{\mathrm{d}\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{t}\right)}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{t}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\int\frac{\mathrm{d}\left(\sqrt{\mathrm{3}}\mathrm{t}−\mathrm{1}\right)}{\sqrt{\mathrm{3}}\mathrm{t}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\mathrm{ln}\mid\frac{\sqrt{\mathrm{3}}\mathrm{t}+\mathrm{1}}{\sqrt{\mathrm{3}}\mathrm{t}−\mathrm{1}}\mid=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\mathrm{ln}\mid\frac{\sqrt{\mathrm{3}\:}\mathrm{arctan}\:\mathrm{x}+\mathrm{1}}{\sqrt{\mathrm{3}}\mathrm{arctan}\:\mathrm{x}−\mathrm{1}}\mid+\mathrm{C} \\ $$

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