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Question Number 48239 by olj55336@awsoo.com last updated on 21/Nov/18

      q.....∫(dx/(sin x cos x+2cos^2 x)), please solve

$$ \\ $$$$ \\ $$$$ \\ $$$${q}.....\int\frac{{dx}}{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}+\mathrm{2cos}\:^{\mathrm{2}} {x}},\:{please}\:{solve} \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 21/Nov/18

I =∫   (dx/((1/(2 ))sin(2x) +1+cos(2x)))dx = ∫   ((2dx)/(sin(2x)+2 +2cos(2x)))  =_(tan(x)=t)       ∫    (2/(((2t)/(1+t^2 )) +2 ((1−t^2 )/(1+t^2 )) +2)) (dt/(1+t^2 ))  = ∫  ((2dt)/(2t +2−2t^2  +2+2t^2 )) =∫  ((2dt)/(2t +4)) =∫ (dt/(t+2)) =ln∣t+2∣ +C  =ln∣2+tan(x)∣ +C .

$${I}\:=\int\:\:\:\frac{{dx}}{\frac{\mathrm{1}}{\mathrm{2}\:}{sin}\left(\mathrm{2}{x}\right)\:+\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{dx}\:=\:\int\:\:\:\frac{\mathrm{2}{dx}}{{sin}\left(\mathrm{2}{x}\right)+\mathrm{2}\:+\mathrm{2}{cos}\left(\mathrm{2}{x}\right)} \\ $$$$=_{{tan}\left({x}\right)={t}} \:\:\:\:\:\:\int\:\:\:\:\frac{\mathrm{2}}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\mathrm{2}\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\mathrm{2}}\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}{t}\:+\mathrm{2}−\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}{t}\:+\mathrm{4}}\:=\int\:\frac{{dt}}{{t}+\mathrm{2}}\:={ln}\mid{t}+\mathrm{2}\mid\:+{C} \\ $$$$={ln}\mid\mathrm{2}+{tan}\left({x}\right)\mid\:+{C}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Nov/18

∫(dx/(tanxcos^2 x+2cos^2 x))  ∫((sec^2 xdx)/(2+tan^ x))  ∫((d(2+tanx))/(2+tanx))  ln(2+tanx)+c

$$\int\frac{{dx}}{{tanxcos}^{\mathrm{2}} {x}+\mathrm{2}{cos}^{\mathrm{2}} {x}} \\ $$$$\int\frac{{sec}^{\mathrm{2}} {xdx}}{\mathrm{2}+{tan}^{} {x}} \\ $$$$\int\frac{{d}\left(\mathrm{2}+{tanx}\right)}{\mathrm{2}+{tanx}} \\ $$$${ln}\left(\mathrm{2}+{tanx}\right)+{c} \\ $$

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