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Question Number 118111 by bemath last updated on 15/Oct/20

∫_0 ^(π/4)  (x^2 /((x sin x+cos x)^2 )) dx =?

π/40x2(xsinx+cosx)2dx=?

Answered by Lordose last updated on 15/Oct/20

  Ω=∫_0 ^( (π/4)) (x^2 /((xsinx+cosx)^2 ))dx  Ω=∫_( 0) ^( (π/4))  ((x^2 sin^2 x+x^2 cos^2 x)/((xsinx+cosx)^2 ))dx  Ω=∫_( 0) ^( (π/4))  ((x^2 sin^2 x+xsinxcosx)/((xsinx+cosx)^2 ))dx + ∫_( 0) ^( (π/4))  ((x^2 cos^2 x−xsinxcosx)/((xsinx+cosx)^2 ))dx  Ω=∫_( 0) ^( (π/4))  ((xsinx(xsinx+cosx))/((xsinx+cosx)^2 ))dx + ∫_( 0) ^( (π/4))  ((−xcosx(−xcosx+sinx))/((xsinx+cosx)^2 ))dx  Ω=Φ + Λ  resolving Λ by IBP  u=sinx−xcosx ⇒ du= xsinxdx  dv=((−xcosx)/((xsinx+cosx)^2 ))dx ⇒ v=(1/(xsinx+cosx))  Λ=∣((sinx−xcosx)/(xsinx+cosx))∣_0 ^(π/4)  − ∫_( 0) ^( (π/4)) (( xsinx)/(xsinx+cosx))dx  Λ=∣((sinx−xcosx)/(xsinx+cosx))∣_0 ^(π/4)  − Φ  Ω= Φ+∣((sinx−xcosx)/(xsinx+cosx))∣_0 ^(π/4) −Φ  Ω=∣((sinx−xcosx)/(xsinx+cosx))∣_0 ^(π/4)   Ω=((4−π)/(4+π))

Ω=0π4x2(xsinx+cosx)2dxΩ=0π4x2sin2x+x2cos2x(xsinx+cosx)2dxΩ=0π4x2sin2x+xsinxcosx(xsinx+cosx)2dx+0π4x2cos2xxsinxcosx(xsinx+cosx)2dxΩ=0π4xsinx(xsinx+cosx)(xsinx+cosx)2dx+0π4xcosx(xcosx+sinx)(xsinx+cosx)2dxΩ=Φ+ΛresolvingΛbyIBPu=sinxxcosxdu=xsinxdxdv=xcosx(xsinx+cosx)2dxv=1xsinx+cosxΛ=∣sinxxcosxxsinx+cosx0π40π4xsinxxsinx+cosxdxΛ=∣sinxxcosxxsinx+cosx0π4ΦΩ=Φ+sinxxcosxxsinx+cosx0π4ΦΩ=∣sinxxcosxxsinx+cosx0π4Ω=4π4+π

Commented by bemath last updated on 15/Oct/20

gave kudos

gavekudos

Answered by Ar Brandon last updated on 15/Oct/20

(d/dx)(xsinx+cosx)=xcosx+sinx−sinx=xcosx  I=∫_0 ^(π/4) (x^2 /((xsinx+cosx)^2 ))dx=∫_0 ^(π/4) ((xcosx)/((xsinx+cosx)^2 ))∙(x/(cosx))dx     ={(x/(cosx))∫((xcosx)/((xsinx+cosx)^2 ))dx}_0 ^(π/4) −∫_0 ^(π/4) {(d/dx)((x/(cosx)))∙∫((xcosx)/((xsinx+cosx)^2 ))dx}     =−[(x/(cosx))∙(1/(xsinx+cosx))]_0 ^(π/4) +∫_0 ^(π/4) {(((cosx+xsinx)/(cos^2 x)))((1/(xsinx+cosx)))}dx     =−[(x/(cosx))∙(1/(xsinx+cosx))]_0 ^(π/4) +∫_0 ^(π/4) (dx/(cos^2 x))     =[tanx−(x/(cosx))∙(1/(xsinx+cosx))]_0 ^(π/4) =[1−((2π)/(4(√2)))∙(8/(π(√2)+4(√2)))]     =1−((16π)/(8π+32))=((8π+32−16π)/(8π+32))=((32−8π)/(32+8π))=((4−π)/(4+π))

ddx(xsinx+cosx)=xcosx+sinxsinx=xcosxI=0π4x2(xsinx+cosx)2dx=0π4xcosx(xsinx+cosx)2xcosxdx={xcosxxcosx(xsinx+cosx)2dx}0π40π4{ddx(xcosx)xcosx(xsinx+cosx)2dx}=[xcosx1xsinx+cosx]0π4+0π4{(cosx+xsinxcos2x)(1xsinx+cosx)}dx=[xcosx1xsinx+cosx]0π4+0π4dxcos2x=[tanxxcosx1xsinx+cosx]0π4=[12π428π2+42]=116π8π+32=8π+3216π8π+32=328π32+8π=4π4+π

Commented by bemath last updated on 15/Oct/20

santuyy

santuyy

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