integration by substitution: (1/x^2) sin(1/x) cos(1/x)

[yossarian]

The problem:

indefinate integral, ((1/x^2)*sin(1/x)*cos(1/x))

X to the negative 2 times the sin of 1/x times the cos of 1/x.

What I have tried:

since f(g(x))*g'(x)*dx=goodies or whatever, let u=g(x)

I set u=sin(1/x), u'=cos(1/x)*(-1/x^2)*dx, dx=-1

So now we have -1 * integral(u*du)

so... the integrant (integred? whatever its called that is differentated to end up with u*u'*dx)

is = to... -1 * (antiderivate(u)) = -cos(u), which doesn't work due to the chain rule.

HALP!

Thanks in advance.[/list]

EDIT: the book says the answer is
(-(sin(1/x))^2) / 2

 

[daon]

\(\displaystyle \L \int sin(\frac{1}{x}) cos(\frac{1}{x})\frac{1}{x^2}dx\)

\(\displaystyle u = sin(\frac{1}{x})\)
\(\displaystyle du = cos(\frac{1}{x}) (-\frac{1}{x^2}) dx\)

\(\displaystyle \L \int sin(\frac{1}{x})cos(\frac{1}{x}) \frac{1}{x^2} dx \,\, = \,\, - \int u du\)