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0-pi-2-cos-2x-ln-sin-x-dx-pi-4-solution-1-0-pi-2-2cos-2-x-1-ln-sin-x-dx-2-0-pi-2-cos-2-x-ln-sin-x-dx-0-pi-2-




Question Number 153555 by mnjuly1970 last updated on 08/Sep/21
       Ω := ∫_0 ^( (π/2)) cos(2x).ln(sin(x))dx=^?  −(π/4)            solution (1 )       Ω := ∫_0 ^( (π/2)) ( 2cos^( 2) (x)−1)ln(sin(x))dx        := 2∫_0 ^( (π/2)) cos^( 2) (x).ln(sin(x))dx−∫_0 ^( (π/2)) ln(sin(x))dx       we know that : ∫_0 ^(π/2) ln(sin(x))dx=_(earlier) ^(derived)  ((−π)/2) ln(2)               ∫_0 ^( (π/2)) cos^( 2) (x).ln(sin(x))dx=_(posts) ^(previous)  −(π/4)ln(2)−(π/8)        ∴   Ω := −(π/2) ln(2) −(π/4) +(π/2) ln(2)                       ◂    Ω =− (π/4)  ▶      m.n
Ω:=0π2cos(2x).ln(sin(x))dx=?π4solution(1)Ω:=0π2(2cos2(x)1)ln(sin(x))dx:=20π2cos2(x).ln(sin(x))dx0π2ln(sin(x))dxweknowthat:0π2ln(sin(x))dx=derivedearlierπ2ln(2)0π2cos2(x).ln(sin(x))dx=previouspostsπ4ln(2)π8Ω:=π2ln(2)π4+π2ln(2)Ω=π4m.n
Answered by Ar Brandon last updated on 08/Sep/21
Ω=∫_0 ^(π/2) cos(2x)ln(sinx)dx   { ((u(x)=ln(sinx))),((v′(x)=cos(2x))) :}⇒ { ((u′(x)=((cosx)/(sinx)))),((v(x)=((sin2x)/2))) :}  Ω=[((ln(sinx)sin2x)/2)−∫cos^2 xdx]_0 ^(π/2)       =−∫_0 ^(π/2) cos^2 xdx=−(1/2)∫_0 ^(π/2) (1+cos2x)dx      =−(1/2)[x+((sin2x)/2)]_0 ^(π/2) =−(π/4)
Ω=0π2cos(2x)ln(sinx)dx{u(x)=ln(sinx)v(x)=cos(2x){u(x)=cosxsinxv(x)=sin2x2Ω=[ln(sinx)sin2x2cos2xdx]0π2=0π2cos2xdx=120π2(1+cos2x)dx=12[x+sin2x2]0π2=π4
Commented by mnjuly1970 last updated on 08/Sep/21
 very nice sir brando as always..
verynicesirbrandoasalways..
Commented by Ar Brandon last updated on 08/Sep/21
Thanks

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