Question 355862
Use a substitution.
Let {{{x=sin(u)}}}, {{{dx=cos(u)du}}}
{{{x^2=(sin(u))^2}}}
{{{1-x^2=1-(sin(u))^2=(cos(u))^2}}}
{{{sqrt(1-x^2)=cos(u)}}}
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{{{ int ( x^2 / ( sqrt (1-(x^2) ) ), dx ) = int ( (sin(u))^2 / ( cos(u) ) , cos(u)du ) }}}
{{{ int ( x^2 / ( sqrt (1-(x^2) ) ), dx ) = int ( (sin(u))^2 , du ) }}}
Take it from there and solve for the integral in u, then back substitute to x.
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Complete the square in the denominator.
{{{ int ( 1/ (sqrt (29-4x+(x^2)) ), dx ) =int ( 1/ (sqrt (x^2-4x+4+25)) ), dx ) }}}
{{{ int ( 1/ (sqrt (29-4x+(x^2)) ), dx ) =int ( 1/ (sqrt ((x-2)^2+25)) ), dx ) }}}
Substitute, let {{{u=x-2}}}, {{{du=dx}}}
{{{ int ( 1/ (sqrt (29-4x+(x^2)) ), dx ) =int ( 1/ (sqrt (u^2+25)) ), du) }}}
Substitute again, let {{{u=5tan(v)}}}, {{{du=5(sec(v)^2)dv}}}
{{{u^2+25=25(tan(v))^2+25=25((tan(v))^2+1)=25(sec(v))^2}}}
{{{sqrt(u^2+25)=5*sec(v)}}}
{{{1/sqrt(u^2+25)=1/(5*sec(v))}}}
{{{ int ( 1/ (sqrt (29-4x+(x^2)) ), dx ) =int ( 1/ (5*sec(v)), 5(sec(v)^2)dv) }}}
{{{ int ( 1/ (sqrt (29-4x+(x^2)) ), dx ) =int ( sec(v), dv) }}}
Take it from here and get the solution in v, back substitute for u, and then back substitute for x.
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