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?