What is the integration of #int((x(tanx)^-1)dx)/(1+x^2)^(3/2)# ?

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#int((x(tanx)^-1)dx)/(1+x^2)^(3/2)#

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Answer 1 · 2018-07-30T12:26:01

#int((x(tanx)^-1)dx)/(1+x^2)^(3/2)=-(tan^-1x)(1/sqrt(1+x^2))+x/sqrt(1+x^2)+C#

#int((x(tanx)^-1)dx)/(1+x^2)^(3/2)#

Put #x=tanz#

#dx=sec^2z#

#int (tanz.z.(sec^2z))/((sec^2z)^(3/2)).dz#

#int(tanz.z.dz)/secz.dz#

#int(sinz/cosz).cosz.z.dz#

#intsinz.z.dz#

#intu.dv = u.v -intdu.v#

#u = z dz #
#du = 1#

#dv=sinz#
#v=-cosz#

Now put values in formula

#intz.sinz.dz=z(-cosz)-int(-cosz)#

#intz.sinz.dz=z(-cosz)+intcosz#

#intz.sinz.dz=z(-cosz)+sinz +C#.......(1)

As we know #cosz=1/secz#

#secz=sqrt(1+tan^2z)#

And is equal to #tanz =x# above.

then, #secz=sqrt(1+x^2)#

#cosz=1/sqrt(1+x^2)#

#sinz=sqrt(1-cos^2z)#

#sinz=sqrt(1-1/(1+x^2))#

#sinz=sqrt(x^2/(1+x^2))#

#sinz = x/sqrt(1+x^2)#

Now put value of z, cosz, sinz in eq 1

Here is Final answer#int((x(tanx)^-1)dx)/(1+x^2)^(3/2)=tan^-1x(-1/sqrt(1+x^2))+x/sqrt(1+x^2) +C#