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Question Number 55930 by ajfour last updated on 06/Mar/19
∫_0 ^( π) ((xtan x)/(sec x+tan x))dx = (is it (π^2 /2)−π)?
∫0πxtanxsecx+tanxdx=(isitπ22−π)?
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19
∫_0 ^π ((xsinx)/(1+sinx))dx I=∫_0 ^π (((π−x)sin(π−x))/(1+sin(π−x)))dx =∫_0 ^π (((π−x)sinx)/(1+sinx))dx 2I=∫_0 ^π ((πsinx−xsinx+xsinx)/(1+sinx))dx 2I=π∫_0 ^π ((sinx)/(1+sinx))dx =π∫_0 ^π (1−(1/(1+sinx)))dx ((2I)/π)=∫_0 ^π dx−∫_0 ^π ((1−sinx)/(cos^2 x))dx ((2I)/π)=∫_0 ^π dx−∫_0 ^π sec^2 xdx+∫_0 ^π secxtanxdx ((2I)/π)=∣x−tanx+secx∣_0 ^π ((2I)/π)=(π−0−1)−(0−0+1) ((2I)/π)=π−1−1 2I=π^2 −2π I=(π^2 /2)−π
∫0πxsinx1+sinxdxI=∫0π(π−x)sin(π−x)1+sin(π−x)dx=∫0π(π−x)sinx1+sinxdx2I=∫0ππsinx−xsinx+xsinx1+sinxdx2I=π∫0πsinx1+sinxdx=π∫0π(1−11+sinx)dx2Iπ=∫0πdx−∫0π1−sinxcos2xdx2Iπ=∫0πdx−∫0πsec2xdx+∫0πsecxtanxdx2Iπ=∣x−tanx+secx∣0π2Iπ=(π−0−1)−(0−0+1)2Iπ=π−1−12I=π2−2πI=π22−π
Commented by ajfour last updated on 06/Mar/19
Thanks Sir, great!
ThanksSir,great!
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19
thank you sir...
thankyousir…
Commented by maxmathsup by imad last updated on 09/Mar/19
I =∫_0 ^π ((x(1+sinx−1))/(1+sinx))dx =∫_0 ^π xdx −∫_0 ^π (x/(1+sinx)) dx but ∫_0 ^π xdx =[(x^2 /2)]_0 ^π =(π^2 /2) and ∫_0 ^π (x/(1+sinx)) dx =_(tan((x/2))=t) ∫_0 ^∞ ((2arctan(t))/(1+((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) =4 ∫_0 ^∞ ((arctan(t))/(1+t^2 +2t)) dt =4 ∫_0 ^∞ ((arctan(t))/((t+1)^2 )) dt ( and by parts) =4{ [−((arctan(t))/(t+1))]_0 ^∞ +∫_0 ^∞ (1/((t+1)(t^2 +1)))dt} =4 ∫_0 ^∞ (dt/((t+1)(t^2 +1))) dt 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→+∞) tF(t) = 0 =a+b ⇒b =−a =−(1/2) ⇒F(t)=(1/(2(t+1))) +((−(t/2)+c)/(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)) ⇒ ∫_0 ^∞ F(t)dt =(1/2)[ln∣t+1∣−(1/2)ln(t^2 +1)]_0 ^(+∞) +(1/2)[arctan(t)]_0 ^(+∞) =(1/2)[ln∣((t+1)/( (√(t^2 +1))))∣]_0 ^∞ +(π/4) =(π/4) ⇒ I =(π^2 /2) −4((π/4)) =(π^2 /2)−π .
I=∫0πx(1+sinx−1)1+sinxdx=∫0πxdx−∫0πx1+sinxdxbut∫0πxdx=[x22]0π=π22and∫0πx1+sinxdx=tan(x2)=t∫0∞2arctan(t)1+2t1+t22dt1+t2=4∫0∞arctan(t)1+t2+2tdt=4∫0∞arctan(t)(t+1)2dt(andbyparts)=4{[−arctan(t)t+1]0∞+∫0∞1(t+1)(t2+1)dt}=4∫0∞dt(t+1)(t2+1)dtletdecomposeF(t)=1(t+1)(t2+1)⇒F(t)=at+1+bt+ct2+1a=limt→−1(t+1)F(t)=12limt→+∞tF(t)=0=a+b⇒b=−a=−12⇒F(t)=12(t+1)+−t2+ct2+1F(0)=1=12+c⇒c=12⇒F(t)=12(t+1)−12t−1t2+1⇒∫0∞F(t)dt=12[ln∣t+1∣−12ln(t2+1)]0+∞+12[arctan(t)]0+∞=12[ln∣t+1t2+1∣]0∞+π4=π4⇒I=π22−4(π4)=π22−π.
Answered by MJS last updated on 06/Mar/19
∫((xtan x)/(sec x +tan x))dx=∫((xsin x)/(1+sin x))dx=∫xdx−∫(x/(1+sin x))dx= =(x^2 /2)−∫(x/(1+sin x))dx= ∫(x/(1+sin x))dx= [∫u′v=uv−∫uv′; u′=(1/(1+sin x)); v=x u=−((cos x)/(1+sin x)); v′=1] =−((xcos x)/(1+sin x))+∫((cos x)/(1+sin x))dx= =−((xcos x)/(1+sin x))+ln (1+sin x) =(x^2 /2)+((xcos x)/(1+sin x))−ln (1+sin x) +C ⇒ your value is true
∫xtanxsecx+tanxdx=∫xsinx1+sinxdx=∫xdx−∫x1+sinxdx==x22−∫x1+sinxdx=∫x1+sinxdx=[∫u′v=uv−∫uv′;u′=11+sinx;v=xu=−cosx1+sinx;v′=1]=−xcosx1+sinx+∫cosx1+sinxdx==−xcosx1+sinx+ln(1+sinx)=x22+xcosx1+sinx−ln(1+sinx)+C⇒yourvalueistrue
Commented by ajfour last updated on 06/Mar/19
Thanks MJS Sir!
ThanksMJSSir!

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