# 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?

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Are you sure these two integrals are equal? Wolframalpha reports $\pi\int f(x)^2 =3\pi/5$ and $\pi\int g(x)^2=2\pi/5$. – matt Jan 1 '12 at 21:40
@N3buchadnezzar Have you received my email? Thanks. – Olivier Oloa Dec 17 '14 at 3:17

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.