How to find the primitive of this guy by parts? I tried:
$$\int \sin(\ln x)dx $$ $$=\int\cos(\ln x)*\tan(\ln x)dx$$ $$=\frac{1}{x*\cos(\ln x)}+\int\frac{1}{x*\cos^2(\ln x)}*\frac{1}{x}\sin(\ln x)dx$$
Is this the way to go?
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If you're not set on integration by parts, there's a cute proof that uses $\sin(\ln x)=\frac{1}{2i}\left(x^i+x^{-i}\right)$, where $i=\sqrt{-1}$. – πr8 Dec 21 '16 at 19:29
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Things may be clearer if you make the change of variable $$ u=\ln x,\quad x=e^u,\quad dx=e^udu, $$ obtaining $$ \int \sin(\ln x)dx=\int e^u\sin(u)du $$ then making the integration by parts.
Olivier Oloa
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$$\int \sin(\ln x)dx=\int x\cdot \frac{1}{x}\sin(\ln x)dx=\int x\cdot(-\cos(\ln x))^{'}dx$$
$$=-x\cos(\ln x)-\int1\cdot(-\cos(\ln x))\,dx=-x\cos(\ln x)+\int\cos(\ln x)\,dx$$
Now,
$$\int \cos(\ln x)dx=\int x\cdot \frac{1}{x}\cos(\ln x)dx=\int x\cdot(\sin(\ln x))^{'}dx$$
$$=x\sin(\ln x)-\int1\cdot\sin(\ln x)\,dx=x\sin(\ln x)-\int\sin(\ln x)\,dx$$
So:
$$\int \sin(\ln x)dx=-x\cos(\ln x)+x\sin(\ln x)-\int \sin(\ln x)dx$$
giving the final integral as $\frac{-x\cos(\ln x)+x\sin(\ln x)}{2}+C$
πr8
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