Primitivation and trigonometry: $\int \sin(\ln x)dx $

Question

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?

Answer 1

$$\int \sin(\ln x)dx=\int x\cdot \frac{1}{x}\sin(\ln x)dx=\int x\cdot(-\cos(\ln x))^{'}dx$$

$$=-x\cos(\ln x)-\int1\cdot(-\cos(\ln x))\,dx=-x\cos(\ln x)+\int\cos(\ln x)\,dx$$

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 $\frac{-x\cos(\ln x)+x\sin(\ln x)}{2}+C$

Answer 2

hint

Use $u=\sin(\ln x)$ and $dv= 1 dx$. Then use integration by parts again.

Answer 3

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.