Evaluate $\int \sin \left(\log x\right) \; dx$
Let $I = \int \sin \left(\log x\right) \; dx$ $\;\;\; \cdots$ (1)
Let $\log x = t$ $\;\;\; \cdots$ (2)
$\implies$ $x = e^t$ $\;\;\; \cdots$ (2a)
Differentiating equation (2) gives
$\dfrac{dx}{x} = dt$ $\implies$ $dx = x \; dt$ $\implies$ $dx = e^t \; dt$ $\;\;\; \cdots$ (2b) $\;\;\;$ [From equation (2a)]
In view of equations (2) and (2b), equation (2a) becomes
$\therefore$ $\;$ $I = \int \sin t \; e^t \; dt$ $\;\;\; \cdots$ (3)
$\left[\text{Note: } \displaystyle \int u \cdot v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \right]$
Here $u = \sin t$ and $v = e^t$
$\begin{aligned}
\therefore \; I & = \sin t \int e^t \; dt - \int \left\{\int e^t \; dt \times \dfrac{d}{dt} \left(\sin t\right) \right\} \; dt \\\\
& = \sin t \cdot e^t - \int e^t \cdot \cos t \; dt \;\;\; \left[\text{Here } u = \cos t \text{ and } v = e^t\right] \\\\
& = e^t \cdot \sin t - \left[\cos t \int e^t \; dt - \int \left\{\int e^t \; dt \times \dfrac{d}{dt} \left(\cos t\right) \right\} \; dt\right] \\\\
& = e^t \cdot \sin t - \left[e^t \cdot \cos t + \int e^t \cdot \sin t \; dt\right] \\\\
& = e^t \cdot \sin t - e^t \cdot \cos t - I + c \;\;\; \left[\text{From equation (3)}\right] \\\\
i.e. \; 2 \; I & = e^t \cdot \sin t - e^t \cdot \cos t + c \\\\
i.e. \; I & = \dfrac{e^t}{2} \left[\sin t - \cos t\right] + c \\\\
& = \dfrac{x}{2} \left[\sin \left(\log x\right) - \cos \left(\log x\right)\right] + c \;\;\; \left[\text{From equations (2) and (2a)}\right]
\end{aligned}$