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I-0-1-x-ln-1-x-2-dx-I-sin-x-cos-3-x-dx-




Question Number 115464 by bemath last updated on 26/Sep/20
I= ∫_(0 ) ^1  x ln (1+x^2 ) dx ?  I=∫ (√(sin x)) .cos^3 x dx ?
I=10xln(1+x2)dx?I=sinx.cos3xdx?
Answered by malwaan last updated on 26/Sep/20
∫(√(sin x)) cos^3 x dx  =∫(√(sin x)) cos^2 x cos x dx  =∫(√(sin x)) (1−sin^2 x)d(sin x)  =∫(√u)(1−u^2 )du  {u=sin x}  y=(√u)⇒y^2 =u⇒2ydy=du  ∴ 2∫y(1−y^4 )ydy=2∫(y^2 −y^6 )dy  = 2((y^3 /3) − (y^6 /6))+ C  = 2(((((√u))^3 )/3) − ((((√u))^6 )/6) + C  = 2(((u(√u))/3) − (u^3 /6)) + C  = 2(((sinx(√(sinx)))/3) − ((sin^3 x)/6))+ C
sinxcos3xdx=sinxcos2xcosxdx=sinx(1sin2x)d(sinx)=u(1u2)du{u=sinx}y=uy2=u2ydy=du2y(1y4)ydy=2(y2y6)dy=2(y33y66)+C=2((u)33(u)66+C=2(uu3u36)+C=2(sinxsinx3sin3x6)+C
Commented by bemath last updated on 26/Sep/20
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Commented by malwaan last updated on 26/Sep/20
gave kudos  What is this ?please
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Commented by bemath last updated on 26/Sep/20
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Commented by bemath last updated on 26/Sep/20
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Answered by malwaan last updated on 26/Sep/20
∫x ln(1+x^2 )dx  u=ln(1+x^2 )⇒du=((2x)/(1+x^2 )) dx  dv = xdx⇒v=(x^2 /2)  ∫udv=uv−∫vdu  ∴∫x ln(1+x^2 )dx=  (x^2 /2)ln(1+x^2 )−∫(x^3 /(1+x^2 )) dx  =(x^2 /2)ln(1+x^2 )−∫[x−(x/(1+x^2 ))]dx  =(x^2 /2)ln(1+x^2 )− (x^2 /2) + (1/2)ln(1+x^2 )+C  ∴  _0 ∫^1 x ln(1+x^2 )dx  = [(1/2)ln(2)− (1/2) +(1/2)ln(2)]− [0]  =ln(2)−(1/2)
xln(1+x2)dxu=ln(1+x2)du=2x1+x2dxdv=xdxv=x22udv=uvvduxln(1+x2)dx=x22ln(1+x2)x31+x2dx=x22ln(1+x2)[xx1+x2]dx=x22ln(1+x2)x22+12ln(1+x2)+C01xln(1+x2)dx=[12ln(2)12+12ln(2)][0]=ln(2)12
Answered by Dwaipayan Shikari last updated on 26/Sep/20
∫_0 ^1 xlog(1+x^2 )dx  =[log(1+x^2 )(x^2 /2)]_0 ^1 −∫_0 ^1 (x^2 /2).((2x)/(1+x^2 ))  =(1/2)log(2)−∫_0 ^1 x−(x/(1+x^2 ))  =(1/2)log(2)−(1/2)+[(1/2)log(1+x^2 )]_0 ^1   =log(2)−(1/2)
01xlog(1+x2)dx=[log(1+x2)x22]0101x22.2x1+x2=12log(2)01xx1+x2=12log(2)12+[12log(1+x2)]01=log(2)12
Answered by Ar Brandon last updated on 26/Sep/20
I=∫_0 ^1 xln(1+x^2 )dx=(1/2)∫_0 ^1 2xln(1+x^2 )dx     =(1/2)∫_0 ^1 ln(1+x^2 )d(x^2 )=[(((1+x^2 )[ln(1+x^2 )−1])/2)]_0 ^1      =(ln2−1)+(1/2)=ln2−(1/2)
I=01xln(1+x2)dx=12012xln(1+x2)dx=1201ln(1+x2)d(x2)=[(1+x2)[ln(1+x2)1]2]01=(ln21)+12=ln212
Answered by mathmax by abdo last updated on 26/Sep/20
I =∫_0 ^1  xln(1+x^2 )dx  by parts we get   I =[(x^2 /2)ln(1+x^2 )]_0 ^1 −∫_0 ^1 (x^2 /2).((2x)/(1+x^2 ))dx =((ln(2))/2)−∫_0 ^1 (x^3 /(x^2 +1))dx but  ∫_0 ^1  (x^3 /(x^2  +1))dx =∫_0 ^1 ((x(x^2 +1)−x)/(x^2  +1))dx =∫_0 ^1 xdx−∫_0 ^1  ((xdx)/(x^2  +1))  =(1/2)−[(1/2)ln(1+x^2 )]_0 ^1 =(1/2)−(1/2)ln(2) ⇒I =((ln(2))/2)−(1/2) +((ln(2))/2)  =ln(2)−(1/2)
I=01xln(1+x2)dxbypartswegetI=[x22ln(1+x2)]0101x22.2x1+x2dx=ln(2)201x3x2+1dxbut01x3x2+1dx=01x(x2+1)xx2+1dx=01xdx01xdxx2+1=12[12ln(1+x2)]01=1212ln(2)I=ln(2)212+ln(2)2=ln(2)12

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