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Question Number 48239 by olj55336@awsoo.com last updated on 21/Nov/18
q.....∫dxsinxcosx+2cos2x,pleasesolve
Commented by maxmathsup by imad last updated on 21/Nov/18
I=∫dx12sin(2x)+1+cos(2x)dx=∫2dxsin(2x)+2+2cos(2x)=tan(x)=t∫22t1+t2+21−t21+t2+2dt1+t2=∫2dt2t+2−2t2+2+2t2=∫2dt2t+4=∫dtt+2=ln∣t+2∣+C=ln∣2+tan(x)∣+C.
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Nov/18
∫dxtanxcos2x+2cos2x∫sec2xdx2+tanx∫d(2+tanx)2+tanxln(2+tanx)+c
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