Solve the given integral \[\int{\sin \left( \log x \right)}dx=\]
A. \[\dfrac{x}{2}\left[ \sin (\log x)-\cos (\log x) \right]\]
B. \[\cos (\log x)-x\]
C. \[\int{\dfrac{(x-1){{e}^{x}}}{{{(x+1)}^{3}}}}\]
D. \[-\cos (\log x)\]

Hint: To solve the above problem we have to use integration by parts formula and it is given by \[\int{u.v=u\int{v}}-\int{du\int{v}}\]. So consider u as \[\sin \left( \log x \right)\] and v as 1 and apply the integration by parts formula as stated above. The derivative of \[\sin t\] is \[\cos t\].

Complete step-by-step solution -
To find the \[\int{\sin \left( \log x \right)}dx\]
Take I as \[\int{\sin \left( \log x \right)}\times 1dx\]
\[I=\int{\sin \left( \log x \right)}\times 1dx\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . (1)
Apply integration by parts formula
\[I\]\[=\sin \left( \log x \right)\times x-\int{\cos (\log x)\times \dfrac{1}{x}}\times xdx\]
\[I\]\[=\sin \left( \log x \right)\times x-\int{\cos (\log x)\times 1}dx\]. . . . . . . . . . . . . . .(2)
\[I\]\[=\sin \left( \log x \right)\times x-\cos (\log x)\times x-\int{\sin \left( \log x \right)dx}\]
By equation(1) we substitute the value of $\int{\sin \left( \log x \right)dx}$ as $I$
\[I=\sin \left( \log x \right)\times x-\cos (\log x)\times x-I\]. . . . . . . . . . . . . . . . . . . . . . . . (3)
\[2I=x\left[ \sin \left( \log x \right)-\cos (\log x) \right]\]
\[I=\dfrac{x}{2}\left[ \sin \left( \log x \right)-\cos (\log x) \right]\]. . . . . . . . . . . . . . . . .. . (4)
The answer is option A.

Note: When they ask us to find the integration of a single function we can use normal integration formulas and if they ask us to find integration of product of two functions then we have to apply the integration by parts. If the two functions are different then we have to use the ILATE rule.