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calculate-0-pi-1-2sinx-3-2cosx-dx-let-A-0-pi-1-2sinx-3-2cosx-dx-changement-tan-x-2-t-give-A-0-1-4t-1-t-2-3-2-1-t-2-1-t-2-2dt-1-t-2-2-0-




Question Number 53112 by maxmathsup by imad last updated on 21/Jan/19
calculate ∫_0 ^π   ((1+2sinx)/(3 +2cosx))dx  let A =∫_0 ^π  ((1+2sinx)/(3 +2cosx))dx  changement tan((x/2))=t give  A =∫_0 ^∞    ((1+((4t)/(1+t^2 )))/(3+2((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =2 ∫_0 ^∞     ((1+t^2  +4t)/((1+t^2 )^2 (((3+3t^2 +2−2t^2 )/(1+t^2 )))))dt  =2 ∫_0 ^∞    ((t^2 +4t +1)/((1+t^2 )(5+t^2 )))dt  let decompose F(t)=((t^2  +4t+1)/((t^2 +1)(t^2  +5)))  F(t)=((at +b)/(t^2  +1)) +((ct +d)/(t^2  +5)) ⇒(at+b)(t^2  +5)+(ct+d)(t^2  +1) =t^2  +4t +1 ⇒  at^3  +5at +bt^2  +5b +ct^3  +ct +dt^2  +d =t^2  +4t +1 ⇒  (a+c)t^3  +(b+d)t^2  +(5a+c)t +5b +d =t^2  +4t +1 ⇒a+c=0 and b+d=1 and  5a+c =4 and 5b+d =1 ⇒c=−a ⇒a=1 ⇒c=−1   we have d=1−b ⇒5b +1−b =1 ⇒b=0 ⇒d=1 ⇒  F(t)=(t/(t^2  +1)) +((−t +1)/(t^2  +5))  ⇒ A =2 ∫_0 ^∞  F(t)dt =∫_0 ^∞  ((2t)/(t^2  +1))dt +∫_0 ^∞  ((−2t +2)/(t^2  +5))dt  =[ln(((t^2  +1)/(t^2  +5)))]_0 ^(+∞)   +2 ∫_0 ^∞   (dt/(t^2  +5)) =ln(5) + 2 ∫_0 ^∞   (dt/(t^2  +5))  but  ∫_0 ^∞   (dt/(t^2  +5))dt =_(t =(√5)u )    ∫_0 ^∞   (((√5)du)/(5(1+u^2 ))) =(1/( (√5))) [artanu]_0 ^(+∞)  =(π/(2(√5))) ⇒  A =ln(5) +(π/(2(√5))) .
calculate0π1+2sinx3+2cosxdxletA=0π1+2sinx3+2cosxdxchangementtan(x2)=tgiveA=01+4t1+t23+21t21+t22dt1+t2=201+t2+4t(1+t2)2(3+3t2+22t21+t2)dt=20t2+4t+1(1+t2)(5+t2)dtletdecomposeF(t)=t2+4t+1(t2+1)(t2+5)F(t)=at+bt2+1+ct+dt2+5(at+b)(t2+5)+(ct+d)(t2+1)=t2+4t+1at3+5at+bt2+5b+ct3+ct+dt2+d=t2+4t+1(a+c)t3+(b+d)t2+(5a+c)t+5b+d=t2+4t+1a+c=0andb+d=1and5a+c=4and5b+d=1c=aa=1c=1wehaved=1b5b+1b=1b=0d=1F(t)=tt2+1+t+1t2+5A=20F(t)dt=02tt2+1dt+02t+2t2+5dt=[ln(t2+1t2+5)]0++20dtt2+5=ln(5)+20dtt2+5but0dtt2+5dt=t=5u05du5(1+u2)=15[artanu]0+=π25A=ln(5)+π25.
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19
∫(1/(3+2cosx))dx+∫((2sinx)/(3+2cosx))dx  ∫((1+tan^2 (x/2))/(3+3tan^2 (x/2)+2−2tan^2 (x/2)))dx−∫((d(3+2cosx))/(3+2cosx))  ∫((sec^2 (x/2))/(5+tan^2 (x/2)))dx−ln(3+2cosx)+c  2∫((d(tan(x/2)))/(5+tan^2 (x/2)))−do  ∣2×(1/( (√5)))tan^(−1) (((tan(x/2))/( (√5))))−ln(3+2cosx)∣_0 ^π   [{(2/( (√5)))tan^(−1) (((tan(π/2))/( (√5))))−ln(3+2cosπ)}−{(2/( (√5)))tan^(−1) (((tan0)/( (√5))))−ln(3+2cos0)}]  =[{(2/( (√5)))((π/2))−ln(3−2)}−{(2/( (√5)))tan^(−1) (0)−ln5}]  =(π/( (√5)))+ln5
13+2cosxdx+2sinx3+2cosxdx1+tan2x23+3tan2x2+22tan2x2dxd(3+2cosx)3+2cosxsec2x25+tan2x2dxln(3+2cosx)+c2d(tanx2)5+tan2x2do2×15tan1(tanx25)ln(3+2cosx)0π[{25tan1(tanπ25)ln(3+2cosπ)}{25tan1(tan05)ln(3+2cos0)}]=[{25(π2)ln(32)}{25tan1(0)ln5}]=π5+ln5

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