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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.....dxsinxcosx+2cos2x,pleasesolve

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=dx12sin(2x)+1+cos(2x)dx=2dxsin(2x)+2+2cos(2x)=tan(x)=t22t1+t2+21t21+t2+2dt1+t2=2dt2t+22t2+2+2t2=2dt2t+4=dtt+2=lnt+2+C=ln2+tan(x)+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

dxtanxcos2x+2cos2xsec2xdx2+tanxd(2+tanx)2+tanxln(2+tanx)+c

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