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Question Number 63410 by aliesam last updated on 03/Jul/19

Commented by mathmax by abdo last updated on 03/Jul/19

2) let I =∫   ((xdx)/(1+sinx))   changement tan((x/2))=t give   I = ∫   ((2arctan(t))/((1+((2t)/(1+t^2 )))(1+t^2 )))dt = 2∫    ((arctan(t))/(1+t^2  +2t)) dt =2 ∫   ((arctan(t))/((t+1)^2 )) dt    =_(by pwrts)     2{   −((arcant)/(t+1)) −∫  −(1/(t+1)) (dt/(1+t^2 ))}  −((2 arctan(t))/(t+1)) + 2 ∫   (dt/((t+1)(t^2  +1)))  let decompose  F(t) =(1/((t+1)(t^2  +1))) ⇒F(t) =(a/(t+1)) +((bt +c)/(t^2  +1))  a =lim_(t→−1) (t+1)F(t) =(1/2)  lim_(t→+∞) t F(t) =0 =a+b ⇒b=−(1/2) ⇒F(t)=(1/(2(t+1))) −(1/2)  ((t −2c)/(t^2  +1))  F(0) =1 =(1/2) +c ⇒c=(1/2) ⇒F(t)=(1/(2(t+1))) −(1/2) ((t−1)/(t^2  +1)) ⇒  ∫  (dt/((t+1)(t^2  +1))) = ∫  (dt/(2(t+1))) −(1/4) ∫  ((2t)/(t^2  +1)) +(1/2) ∫   (dt/(t^2  +1))  =(1/2)ln∣t+1∣−(1/4)ln(t^2  +1)+(1/2) arctan(t)+c  =(1/2)ln∣1+tan((x/2))∣−(1/4)ln(1+tan^2 ((x/2))) +x +c  ⇒  I =−(x/(1+tan((x/2)))) +ln∣1+tan((x/2))∣−(1/2)ln(1+tan^2 ((x/2))) +x +c .

2)letI=xdx1+sinxchangementtan(x2)=tgiveI=2arctan(t)(1+2t1+t2)(1+t2)dt=2arctan(t)1+t2+2tdt=2arctan(t)(t+1)2dt=bypwrts2{arcantt+11t+1dt1+t2}2arctan(t)t+1+2dt(t+1)(t2+1)letdecomposeF(t)=1(t+1)(t2+1)F(t)=at+1+bt+ct2+1a=limt1(t+1)F(t)=12limt+tF(t)=0=a+bb=12F(t)=12(t+1)12t2ct2+1F(0)=1=12+cc=12F(t)=12(t+1)12t1t2+1dt(t+1)(t2+1)=dt2(t+1)142tt2+1+12dtt2+1=12lnt+114ln(t2+1)+12arctan(t)+c=12ln1+tan(x2)14ln(1+tan2(x2))+x+cI=x1+tan(x2)+ln1+tan(x2)12ln(1+tan2(x2))+x+c.

Commented by aliesam last updated on 04/Jul/19

perfect

perfect

Answered by MJS last updated on 04/Jul/19

2)  ∫(x/(1+sin x))dx  by parts  u′=(1/(1+sin x)) → u=∫(dx/(1+sin x))  v=x → v′=1  ∫u′v=uv−∫uv′  ∫(x/(1+sin x))dx=x∫(dx/(1+sin x))−∫(∫(dx/(1+sin x)))dx  ∫(dx/(1+sin x))=       [t=tan (x/2) → dx=2cos^2  (x/2) dt]  =2∫(dt/((t+1)^2 ))=−(2/(t+1))=−2((cos (x/2))/(cos (x/2) +sin (x/2)))=  =tan x −(1/(cos x))−1  ∫(∫(dx/(1+sin x)))dx=∫tan x dx−∫(dx/(cos x))−∫dx=  =−ln cos x −ln ((cos x)/(1−sin x)) −x=−x+ln (1/(1+sin x))  ⇒  ∫(x/(1+sin x))dx=xtan x −(x/(cos x))−x+x−ln (1/(1+sin x)) =  =xtan x −(x/(cos x))−ln (1/(1+sin x)) +C

2)x1+sinxdxbypartsu=11+sinxu=dx1+sinxv=xv=1uv=uvuvx1+sinxdx=xdx1+sinx(dx1+sinx)dxdx1+sinx=[t=tanx2dx=2cos2x2dt]=2dt(t+1)2=2t+1=2cosx2cosx2+sinx2==tanx1cosx1(dx1+sinx)dx=tanxdxdxcosxdx==lncosxlncosx1sinxx=x+ln11+sinxx1+sinxdx=xtanxxcosxx+xln11+sinx==xtanxxcosxln11+sinx+C

Answered by MJS last updated on 04/Jul/19

1)  ∫(dx/(√(1+tan x)))=       [t=(√(1+tan x)) → dx=((2t)/(t^4 −2t^2 +2))dt]  =2∫(dt/(t^4 −2t^2 +2))=2∫(dt/((t^2 −(√(2+2(√2)))t+(√2))(t^2 +(√(2+2(√2)))t+(√2))))=  (√(2+2(√2)))=a  (√2)=b  =2∫(dt/((t^2 −at+b)(t^2 +at+b)))=  =−(1/(ab))∫((t−a)/(t^2 −at+b))dt+(1/(ab))∫((t+a)/(t^2 +at+b))dt=  =−(1/(2ab))∫((2t−a)/(t^2 −at+b))dt+(1/(2b))∫(dt/(t^2 −at+b))+(1/(2ab))∫((2t+a)/(t^2 +at+b))dt+(1/(2b))∫(dt/(t^2 +at+b))=  =−(1/(2ab))ln (t^2 −at+b) +(1/(b(√(4b−a^2 ))))arctan ((2t−a)/(√(4b−a^2 ))) +(1/(2ab))ln (t^2 +at+b) +(1/(b(√(4b−a^2 ))))arctan ((2t+a)/(√(4b−a^2 ))) =  =(1/(2ab))ln ((t^2 +at+b)/(t^2 −at+b)) +(1/(b(√(4b−a^2 ))))(arctan ((2t−a)/(√(4b−a^2 ))) +arctan ((2t+a)/(√(4b−a^2 ))))  ...

1)dx1+tanx=[t=1+tanxdx=2tt42t2+2dt]=2dtt42t2+2=2dt(t22+22t+2)(t2+2+22t+2)=2+22=a2=b=2dt(t2at+b)(t2+at+b)==1abtat2at+bdt+1abt+at2+at+bdt==12ab2tat2at+bdt+12bdtt2at+b+12ab2t+at2+at+bdt+12bdtt2+at+b==12abln(t2at+b)+1b4ba2arctan2ta4ba2+12abln(t2+at+b)+1b4ba2arctan2t+a4ba2==12ablnt2+at+bt2at+b+1b4ba2(arctan2ta4ba2+arctan2t+a4ba2)...

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