Composite Integration

We have $\displaystyle I=\int \sin(\ln x) ~dx$

Substituting $\ln x=u \implies x=e^u$ and $dx=x~du=e^u~du$

Therefore $I=e^u\sin u~du$ .

Applying integration by parts twice on $I$ (first keep $\sin u$ as the first function and $e^u$ as the second function and second time do the reverse), and adding the two, we get $I=\dfrac{e^u(\sin u- \cos u)}{2}+C=\dfrac{x(\sin(\ln x)-\cos(\ln x))}{2}+C=\dfrac{-x\sin(\frac{\pi}{4}-\ln x)}{\sqrt{2}}+C$ .

So we get $a=-1,~b=4,~c=2$ , therefore $b^a-c=\boxed{-1.75}$ .