Show that two integrals are equal I found these two functions to be rather interesting. 
$$ f(x) = \sin( \ln x) \qquad \text{and} \qquad g(x) = \sin( \ln x ) + \cos( \ln x ) $$
I want to show that when rotating these two functions, bounded by the lines $x=0$ and $x=1$, around the x-axis, the respective volumes of the solids obtained are equal.
This problem can be rewritten as showing that
$$ \pi \int_{0}^{1} \left[ \sin(\ln x ) \right]^2 dx \, = \, \pi \int_{0}^{1} \left[ \sin(\ln x ) + \cos(\ln x) \right]^2 dx  $$
I know that both of these integrals equal $\cfrac{3}{5}\pi$, but I want to show that these two are equal without directly computing them. I tried showing that
$$ \pi \int_{0}^{1} \left[ \sin(\ln x ) \right]^2 \, - \, \left[ \sin(\ln x ) + \cos(\ln x) \right]^2  dx  = 0$$
$$ - \int_{0}^{1} \cos(\ln x) + \sin \left( \ln ( x^2 ) \right)  dx  = 0$$
but there I became stuck. Any help showing that these two integrals are in fact the same?
 A: A simple change of variables $y = -\ln(x)$ makes them into
$$ \begin{eqnarray}
   \int_0^1 f(x)^2 \mathrm{d} x &=& \int_0^\infty \sin^2(y) \mathrm{e}^{-y} \mathrm{d} y = \int_0^\infty \frac{1-\cos(2y)}{2} \mathrm{e}^{-y} \mathrm{d} y \\
   \int_0^1 g(x)^2 \mathrm{d} x &=& \int_0^\infty (\cos(y) - \sin(y))^2 \mathrm{e}^{-y} \mathrm{d} y = \int_0^\infty \left( 1- \sin(2y) \right) \mathrm{e}^{-y} \mathrm{d} y
 \end{eqnarray}
$$
Now, since $\cos(2y) = \operatorname{Re}\left( \mathrm{e}^{2 i y} \right)$ and $\sin(2y) = \operatorname{Im}\left( \mathrm{e}^{2 i y} \right)$:
$$
 \begin{eqnarray}
  \int_0^1 f(x)^2 \mathrm{d} x &=& \frac{1}{2}\left(1 - \operatorname{Re}\left( \frac{1}{1-2 i} \right)\right) = \frac{2}{5} \\
  \int_0^1 g(x)^2 \mathrm{d} x &=& 1 - \operatorname{Im}\left( \frac{1}{1-2 i} \right) = 1-\frac{2}{5} = \frac{3}{5}
 \end{eqnarray}
$$ 
where $\int_0^\infty \mathrm{e}^{-\lambda y} \mathrm{d} y = \frac{1}{\lambda}$ for $\operatorname{Re}(\lambda)>0$ was repeatedly used.
Thus these integrals are not the same.
