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Question Number 76716 by peter frank last updated on 29/Dec/19
prove that  ∫_0 ^π ((xsin x)/(1+cos^2 x))=(π^2 /4)
provethat0πxsinx1+cos2x=π24
Commented by abdomathmax last updated on 30/Dec/19
let I =∫_0 ^π  ((xsinx)/(1+cos^2 x))dx changement x=π−t give  I =−∫_0 ^π (((π−t)sint)/(1+cos^2 t))(−dt)=π ∫_0 ^π ((sint)/(1+cos^2 t))dt−I ⇒  2I =π ∫_0 ^π  ((sint)/(1+cos^2 t))dt  we have  ∫_0 ^π  ((sint)/(1+cos^2 t))dt =_(tan((t/2))=u)   ∫_0 ^∞   (((2u)/(1+u^2 ))/(1+(((1−u^2 )/(1+u^2 )))^2 )) ((2du)/(1+u^2 ))  =4 ∫_0 ^∞    (u/((1+u^2 )^2 (1+(((1−u^2 )^2 )/((1+u^2 )^2 )))))du  =4 ∫_0 ^∞   ((udu)/((1+u^2 )^2  +(1−u^2 )^2 ))  =4 ∫_0 ^∞   ((u du)/(u^4  +2u^2  +1 +u^4 −2u^2  +1))  =4 ∫_0 ^∞   ((udu)/(2u^4  +2)) =2 ∫_0 ^∞   ((udu)/(u^4  +1))  =_(u=z^(1/4) )    2 ∫_0 ^∞ ((z^(1/4)  )/(1+z))×(1/4)z^((1/4)−1) dz  =(1/2)∫_0 ^∞  (z^((1/2)−1) /(1+z))dz =(1/2)×(π/(sin((π/2)))) =(π/2) ⇒  2I =(π^2 /2) ⇒ ★I =(π^2 /4)★
letI=0πxsinx1+cos2xdxchangementx=πtgiveI=0π(πt)sint1+cos2t(dt)=π0πsint1+cos2tdtI2I=π0πsint1+cos2tdtwehave0πsint1+cos2tdt=tan(t2)=u02u1+u21+(1u21+u2)22du1+u2=40u(1+u2)2(1+(1u2)2(1+u2)2)du=40udu(1+u2)2+(1u2)2=40uduu4+2u2+1+u42u2+1=40udu2u4+2=20uduu4+1=u=z1420z141+z×14z141dz=120z1211+zdz=12×πsin(π2)=π22I=π22I=π24
Commented by peter frank last updated on 30/Dec/19
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thankyou
Commented by Tony Lin last updated on 30/Dec/19
∫_0 ^π ((xsinx)/(1+cos^2 x))dx  let π−θ=x, dθ=−dx  ∫_0 ^π xf(sinx)dx  =∫_π ^0 (π−θ)f[sin(π−θ)](−dθ)  =∫_0 ^π (π−θ)f[sinθ]dθ  =π∫_0 ^π f(sinθ)dθ−∫_0 ^π θf(sinθ)dθ  =π∫_0 ^π f(sinx)dx−∫_0 ^π xf(sinx)dx  ⇒∫_0 ^π xf(sinx)dx=(π/2)∫_0 ^π f(sinx)dx  ∫_0 ^π ((xsinx)/(1+cos^2 x))dx  =(π/2)∫_0 ^π ((sinx)/(1+cos^2 x))dx  let cosx=t, dt=−sinxdx  (π/2)∫_(−1) ^1 (1/(1+t^2 ))dt  =πtan^(−1) t∣_0 ^1   =(π^2 /4)
0πxsinx1+cos2xdxletπθ=x,dθ=dx0πxf(sinx)dx=π0(πθ)f[sin(πθ)](dθ)=0π(πθ)f[sinθ]dθ=π0πf(sinθ)dθ0πθf(sinθ)dθ=π0πf(sinx)dx0πxf(sinx)dx0πxf(sinx)dx=π20πf(sinx)dx0πxsinx1+cos2xdx=π20πsinx1+cos2xdxletcosx=t,dt=sinxdxπ21111+t2dt=πtan1t01=π24
Commented by peter frank last updated on 30/Dec/19
thank you
thankyou
Answered by Kunal12588 last updated on 30/Dec/19
I=∫_0 ^π ((x sin x)/(1+cos^2  x))dx  ⇒I=∫_0 ^π (((π−x) sin (π−x))/(1+cos^2  (π−x)))dx=∫_0 ^π (((π−x)sin x)/(1+cos^2  x))dx                         [using ∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx]      adding red ones  2I= π∫_0 ^π ((sin x)/(1+cos^2  x))dx                       (((t=cos x⇒dt=−sin x dx)),((x→0⇒t→1 & x→π⇒t→−1)) )  2I=−π∫_1 ^(−1) (dt/(1+t^2 ))=π∫_(−1) ^1 (dt/(1+t^2 ))=2π∫_0 ^1 (dt/(1+t^2 ))                         [using −∫_b ^a f(x)dx=∫_a ^b f(x)dx]  [and  ∫_(−a) ^(         a) f(x) dx =   { ((∫_0 ^( 2a) f(x) dx ; if f(x) is even function)),((0 ; if f(x) is odd function)) :}]  ⇒I=π[tan^(−1) t]_0 ^1 =π[(π/4)−0]=(π^2 /4)  ∴∫_0 ^π ((x sin x)/(1+cos^2 x))dx=(π^2 /4)     QED
I=0πxsinx1+cos2xdxI=0π(πx)sin(πx)1+cos2(πx)dx=0π(πx)sinx1+cos2xdx[usingabf(x)dx=abf(a+bx)dx]addingredones2I=π0πsinx1+cos2xdx(t=cosxdt=sinxdxx0t1&xπt1)2I=π11dt1+t2=π11dt1+t2=2π01dt1+t2[usingbaf(x)dx=abf(x)dx][andaaf(x)dx={02af(x)dx;iff(x)isevenfunction0;iff(x)isoddfunction]I=π[tan1t]01=π[π40]=π240πxsinx1+cos2xdx=π24QED
Commented by peter frank last updated on 30/Dec/19
thank you
thankyou

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