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Question Number 83226 by 09658867628 last updated on 28/Feb/20
∫101x2+2xcosα+1dx=
Commented by mathmax by abdo last updated on 29/Feb/20
I=∫01dx(x+cosα)2+sin2αchangementx+cosα=sinα)ugiveI=∫cotanα1+cosαsinαsinαdusin2α(1+u2)=1sinα[arctanu]1tanα1tan(α2)=1sinα{arctan(1tan(α2))−arctan(1tanα)}iftanα>0wegetI=1sinα{π2−α2−π2+α}=α2sinα....
Answered by MJS last updated on 28/Feb/20
∫dxx2+2xcosα+1==∫dx(x+cosα)2+1−cos2α=[t=x+cosα1−cos2α→dx=dt1−cos2α][ofcourse1−cos2α=∣sinα∣]=1∣sinα∣∫dtt2+1=arctant∣sinα∣=arctanx+cosα∣sinα∣∣sinα∣+C
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