$If \int \frac{sin (ln (x))}{x} d x = f (x), then the value of - f (1) is (take the constant of integration as 0)$
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Solution

We know the integral of $sin x$ and $ln x$ but not of $sin (ln x)$ . Also we can see the derivative of $ln x$ in the integrand.
So we assume $t = lnx$
$\Rightarrow dt = \frac{dx}{x}$
Thus, our integral becomes
$\int sin t dt = - cos (t) + C$
Also we know that the constant of integration is taken as $0$
Thus substituting back $t$ and taking $C$ as $0$ .
We get $f (x) = - cos (lnx)$
$\Rightarrow - f (1) = cos (ln 1) = cos 0 = 1$

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