How do you integrate #\int \frac { 8\ln 6x \sin ^ { - 1} ( \ln 6x ) } { x } d x#?

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Answer 1 · 2017-11-09T16:59:40

#int [8ln6x*sin^(-1) (Ln6x)*dx]/x#

#=[4(Ln6x)^2-2]*sin^(-1) (Ln6x)+2Ln6x*sqrt[1-(Ln6x)^2]+C#

#int [8ln6x*sin^(-1) (Ln6x)*dx]/x#

After using #u=Ln6x# and #du=(dx)/x# transforms, this integral became

#int 8u*sin^(-1) (u)*du#

After using #z=sin^(-1) (u)#, #u=sinz# and #du=cosz*dz# transforms, this integral became

#int 8sinz*z*cosz*dz#

#=int 4z*sin2z*dz#

#=-2z*cos2z#-#int -2cos2z*dz#

#=-2z*cos2z#+#int 2cos2z*dz#

#=-2z*cos2z+sin2z+C#

#=-2z*[1-2(sinz)^2]+2sinz*cosz+C##=-2sin^(-1) (u)*(1-2u^2)+2u*sqrt(1-u^2)+C#

#=(4u^2-2)*sin^(-1) (u)+2u*sqrt(1-u^2)+C#

#=[4(Ln6x)^2-2]*sin^(-1) (Ln6x)+2Ln6x*sqrt[1-(Ln6x)^2]+C#