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Question Number 43337 by math khazana by abdo last updated on 09/Sep/18

let f(x) =∫_0 ^x   (t/(1+sint))dt  1)find a explicit form of f(x)  2) calculate ∫_0 ^∞    (t/(1+sint)) dt

letf(x)=0xt1+sintdt1)findaexplicitformoff(x)2)calculate0t1+sintdt

Commented by maxmathsup by imad last updated on 11/Sep/18

1) changement  tan((t/2))=u give f(x)= ∫_0 ^(tan((x/2)))    ((2 arctanu)/(1+ ((2u)/(1+u^2 )))) ((2du)/(1+u^2 ))  = ∫_0 ^(tan((x/2)))    ((4 arctanu)/(1+u^2  +2u))du =4∫_0 ^(tan((x/2)))   (( arctanu)/((u+1)^2 )) du    by parts we get  f(x) =4{ [−(1/(1+u)) arctanu]_0 ^(tan((x/2)))   +∫_0 ^(tan((x/2)))    (du/((u+1)(1+u^2 ))}  =−4     (x/(2(1+tan((x/2)))))  +4  ∫_0 ^(tan((x/2)))    (du/((u+1)(u^2  +1))) but  ∫_0 ^(tan((x/2)))     (du/((u+1)(u^2  +1))) =(1/2) ∫_0 ^(tan((x/2))) { (1/(u+1)) −((u−1)/(u^2 +1))}du  =(1/2)[ln∣u+1∣]_0 ^(tan((x/2)))   −(1/4) [ln(u^2  +1)]_0 ^(tan((x/2)))   +(1/2) [ arctanu]_0 ^(tan((x/2)))   =(1/2)ln∣1+tan((x/2))∣ −(1/4)ln∣1+tan^2 ((x/2))∣ +(x/4) ⇒  f(x) =((−2x)/(1+tan((x/2))))  +(1/2)ln∣1+tan((x/2))∣ −(1/4)ln∣1+tan^2 ((x/2))∣ +(x/4) .

1)changementtan(t2)=ugivef(x)=0tan(x2)2arctanu1+2u1+u22du1+u2=0tan(x2)4arctanu1+u2+2udu=40tan(x2)arctanu(u+1)2dubypartswegetf(x)=4{[11+uarctanu]0tan(x2)+0tan(x2)du(u+1)(1+u2}=4x2(1+tan(x2))+40tan(x2)du(u+1)(u2+1)but0tan(x2)du(u+1)(u2+1)=120tan(x2){1u+1u1u2+1}du=12[lnu+1]0tan(x2)14[ln(u2+1)]0tan(x2)+12[arctanu]0tan(x2)=12ln1+tan(x2)14ln1+tan2(x2)+x4f(x)=2x1+tan(x2)+12ln1+tan(x2)14ln1+tan2(x2)+x4.

Commented by maxmathsup by imad last updated on 11/Sep/18

2) the Q is calculate  ∫_0 ^(π/2)    (t/(1+sint))dt

2)theQiscalculate0π2t1+sintdt

Commented by maxmathsup by imad last updated on 11/Sep/18

∫_0 ^(π/2)     (t/(1+sint)) dt  =f((π/2)) =((−π)/(1+1)) +(1/2)ln(2) −(1/4)ln(2) +(π/4)  =−(π/4) +((ln(2))/2)  .

0π2t1+sintdt=f(π2)=π1+1+12ln(2)14ln(2)+π4=π4+ln(2)2.

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