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Question Number 59177 by 2772639291927 last updated on 05/May/19

∫ ((cos 2x)/(cos x)) dx =

$$\int\:\:\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{cos}\:{x}}\:{dx}\:= \\ $$

Commented by mathsolverby Abdo last updated on 05/May/19

I =∫((2cos^2 x−1)/(cosx))dx =∫2cosx dx−∫(dx/(cosx)) but ∫ 2cosxdx =2sinx +c_1 ∫ (dx/(cosx)) =_(tan((x/2))=t) ∫ ((2dt)/((1+t^2 )((1−t^2 )/(1+t^2 )))) =2∫ (dt/(1−t^2 )) =∫ ((1/(1−t)) +(1/(1+t)))dt =ln∣((1+t)/(1−t))∣ +c_2 =ln∣((1+tan((x/2)))/(1−tan((x/2))))∣ +c_2 =ln∣tan((x/2) +(π/4))∣ +c_2 ⇒ ∫ ((cos(2x))/(cosx)) dx = 2sinx +ln∣tan((x/2)+(π/4))∣ +C

$${I}\:=\int\frac{\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}}{{cosx}}{dx}\:=\int\mathrm{2}{cosx}\:{dx}−\int\frac{{dx}}{{cosx}} \\ $$$${but}\:\int\:\mathrm{2}{cosxdx}\:=\mathrm{2}{sinx}\:+{c}_{\mathrm{1}} \\ $$$$\int\:\:\frac{{dx}}{{cosx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$=\mathrm{2}\int\:\:\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:=\int\:\:\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\:+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt} \\ $$$$={ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\:+{c}_{\mathrm{2}} \:={ln}\mid\frac{\mathrm{1}+{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\mid\:+{c}_{\mathrm{2}} \\ $$$$={ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{4}}\right)\mid\:+{c}_{\mathrm{2}} \:\:\Rightarrow \\ $$$$\int\:\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{{cosx}}\:{dx}\:=\:\mathrm{2}{sinx}\:+{ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mid\:+{C} \\ $$

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