How do you integrate $(\frac{1}{x^{2}}) (sin (2 ln x)) d x$ $(1/x^2)(sin(2lnx)) dx$ ?

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Answer 1 · 2015-04-11T14:46:49

$f (x) = \int \frac{1}{x^{2}} \cdot sin (2 ln (x)) d x$ $f(x) = int1/x^2*sin(2ln(x))dx$

$= \frac{1}{2} \int \frac{2}{x^{2}} \cdot sin (2 ln (x)) d x$ $= 1/2int2/x^2*sin(2ln(x))dx$

$u = 2 ln (x)$ $u = 2ln(x)$ note $u = ln (x^{2})$ $u = ln(x^2)$
$d u = \frac{2}{x} d x$ $du = 2/x dx$

$= \frac{1}{2} \int \frac{2}{x} \cdot \frac{1}{x} \cdot sin (2 ln (x)) d x$ $= 1/2int2/x*1/x*sin(2ln(x))dx$

$= \frac{1}{2} \int \frac{1}{x} \cdot sin (u) d u$ $=1/2int1/x*sin(u) du$

$e^{u} = x^{2}$ $e^u = x^2$
$e^{\frac{1}{2} u} = x$ $e^(1/2u)=x$
$e^{- \frac{1}{2} u} = \frac{1}{x}$ $e^(-1/2u) = 1/x$

$f (x) = \int \frac{1}{2} e^{- \frac{1}{2} u} \cdot sin (u) d u$ $f(x)=int1/2e^(-1/2u)*sin(u) du$

By part :

$g (x) = \frac{1}{2} e^{- \frac{1}{2} u}$ $g(x) =1/2e^(-1/2u)$
$g'(x)=-1/4e^(-1/2u)$

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

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

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 (1x2)(sin(2lnx))dx(1/x^2)(sin(2lnx)) dx ?