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cos2x-ln-1-tanx-dx-




Question Number 18224 by Arnab Maiti last updated on 17/Jul/17
∫cos2x ln(1+tanx)dx
$$\int\mathrm{cos2x}\:\mathrm{ln}\left(\mathrm{1}+\mathrm{tanx}\right)\mathrm{dx} \\ $$
Answered by alex041103 last updated on 17/Jul/17
We integrate by parts  u=ln(1+tan x)   dv=cos2x  du=((sec^2 x)/(1+tanx))    v=(1/2)sin2x=sinx cosx  ∫cos2x ln(1+tanx)dx=  =sin x cos x ln(1+tanx)−∫  ((tan x)/(1+tan x))dx  But   ((tan x)/(1+tan x))=1−(1/(1+tan x))=1−((cos x)/(sin x+cos x))=  =1−(1/2)(((cos x +sin x )/(cos x +sin x))+((cos x − sin x)/(cos x + sin x)))  ⇒∫ ((tan x)/(1+tan x))dx=x−(1/2)(x+ln∣cosx + sin x∣)  =(x/2)−((ln∣cos x + sin x∣)/2)  ⇒∫cos2x ln(1+tanx)dx=  =sin x cos x ln(1+tanx)−(x/2)+((ln∣sin x + cos x∣)/2) + C
$${We}\:{integrate}\:{by}\:{parts} \\ $$$${u}={ln}\left(\mathrm{1}+{tan}\:{x}\right)\:\:\:{dv}={cos}\mathrm{2}{x} \\ $$$${du}=\frac{{sec}^{\mathrm{2}} {x}}{\mathrm{1}+{tanx}}\:\:\:\:{v}=\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{x}={sinx}\:{cosx} \\ $$$$\int\mathrm{cos2x}\:\mathrm{ln}\left(\mathrm{1}+\mathrm{tanx}\right)\mathrm{dx}= \\ $$$$={sin}\:{x}\:{cos}\:{x}\:{ln}\left(\mathrm{1}+{tanx}\right)−\int\:\:\frac{{tan}\:{x}}{\mathrm{1}+{tan}\:{x}}{dx} \\ $$$${But}\: \\ $$$$\frac{{tan}\:{x}}{\mathrm{1}+{tan}\:{x}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{tan}\:{x}}=\mathrm{1}−\frac{{cos}\:{x}}{{sin}\:{x}+{cos}\:{x}}= \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{cos}\:{x}\:+{sin}\:{x}\:}{{cos}\:{x}\:+{sin}\:{x}}+\frac{{cos}\:{x}\:−\:{sin}\:{x}}{{cos}\:{x}\:+\:{sin}\:{x}}\right) \\ $$$$\Rightarrow\int\:\frac{{tan}\:{x}}{\mathrm{1}+{tan}\:{x}}{dx}={x}−\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{ln}\mid{cosx}\:+\:{sin}\:{x}\mid\right) \\ $$$$=\frac{{x}}{\mathrm{2}}−\frac{{ln}\mid{cos}\:{x}\:+\:{sin}\:{x}\mid}{\mathrm{2}} \\ $$$$\Rightarrow\int\mathrm{cos2x}\:\mathrm{ln}\left(\mathrm{1}+\mathrm{tanx}\right)\mathrm{dx}= \\ $$$$={sin}\:{x}\:{cos}\:{x}\:{ln}\left(\mathrm{1}+{tanx}\right)−\frac{{x}}{\mathrm{2}}+\frac{{ln}\mid{sin}\:{x}\:+\:{cos}\:{x}\mid}{\mathrm{2}}\:+\:{C} \\ $$$$ \\ $$$$ \\ $$

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