How do you integrate $(1/x^2)(sin(2lnx)) dx$ ?

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Answer 1 · 2015-04-11T14:46:49

$f(x) = int1/x^2*sin(2ln(x))dx$

$= 1/2int2/x^2*sin(2ln(x))dx$

$u = 2ln(x)$ note $u = ln(x^2)$
$du = 2/x dx$

$= 1/2int2/x*1/x*sin(2ln(x))dx$

$=1/2int1/x*sin(u) du$

$e^u = x^2$
$e^(1/2u)=x$
$e^(-1/2u) = 1/x$

$f(x)=int1/2e^(-1/2u)*sin(u) du$

By part :

$g(x) =1/2e^(-1/2u)$
$g'(x)=-1/4e^(-1/2u)$

$h'(x)=sin(u)$
$h(x)=-cos(u)$

$int1/2e^(-1/2u)*sin(u) du=-[1/2e^(-1/2u)cos(u)]-int1/4e^(-1/2u)cos(u)du$

by part again :

$g(x) = 1/4e^(-1/2u)$
$g'(x) = -1/8e^(-1/2u)$

$h'(x)=cos(u)$
$h(x)=sin(u)$

$=int1/2e^(-1/2u)*sin(u)du=-[1/2e^(-1/2u)cos(u)]-([1/4e^(-1/2u)sin(u)]+1/4int1/2e^(-1/2u)sin(u)du)$

$=int1/2e^(-1/2u)*sin(u) du=-[1/2e^(-1/2u)cos(u)]-[1/4e^(-1/2u)sin(u)]-1/4int1/2e^(-1/2u)sin(u)du$

$=4int1/2e^(-1/2u)*sin(u) du=-4[1/2e^(-1/2u)cos(u)]-4[1/4e^(-1/2u)sin(u)]-int1/2e^(-1/2u)sin(u)du$

$=4int1/2e^(-1/2u)*sin(u) du=-2[e^(-1/2u)cos(u)]-[e^(-1/2u)sin(u)]-int1/2e^(-1/2u)sin(u)du$

$=5int1/2e^(-1/2u)*sin(u) du=-2[e^(-1/2u)cos(u)]-[e^(-1/2u)sin(u)]$

$=int1/2e^(-1/2u)*sin(u) du=1/5(-2[e^(-1/2u)cos(u)]-[e^(-1/2u)sin(u)])$

Substitute back : $e^(-1/2u) = 1/x$

So we have :

$=(-1/(5x)(2cos(u)+sin(u)))+C$

How do you integrate (1/x^2)(sin(2lnx)) dx (1/x^2)(sin(2lnx)) dx ?