Stosujemy całkowanie przez części: \[ \begin{split} \int \sin \left ( \ln x \right )dx&=\int(x)' \sin \left ( \ln x \right )dx=\\[6pt] &=x \sin \left ( \ln x \right )-\int x \left ( \sin \left ( \ln x \right ) \right )'dx=\\[6pt] &=x \sin ( \ln x )-\int x \frac{\cos (\ln x)}{x}dx=\\[6pt] &=x \sin ( \ln x )-\int \cos (\ln x)\ dx=\\[6pt] &=x \sin ( \ln x )-\int (x)'\cos (\ln x)\ dx=\\[6pt] &=x \sin ( \ln x )-\left ( x\cos (\ln x) -\int x \frac{-\sin(\ln x)}{x}dx \right )=\\[6pt] &=x \sin ( \ln x )- x\cos (\ln x) -\int \sin(\ln x)\ dx \end{split} \] Zatem mamy równość: \[ \begin{split} \int \sin \left ( \ln x \right )dx&=x \sin ( \ln x )- x\cos (\ln x) -\int \sin(\ln x)\ dx\\[6pt] \int \sin \left ( \ln x \right )dx+\int \sin(\ln x)\ dx&=x \sin ( \ln x )- x\cos (\ln x)\\[6pt] 2\int \sin \left ( \ln x \right )dx&=x \sin ( \ln x )- x\cos (\ln x)\\[6pt] \int \sin \left ( \ln x \right )dx&=\frac{1}{2}x \sin ( \ln x )- \frac{1}{2}x\cos (\ln x)+C \end{split} \]