Question

(intsin|ln x|dx)

Explanation:

Step1: Use substitution

Let $u = \ln x$, then $x=e^{u}$ and $dx = e^{u}du$. The integral becomes $\int\sin|u|e^{u}du$. We consider two cases based on the sign of $u$.

Case 1: When $u\geq0$

The integral is $\int\sin(u)e^{u}du$. Use integration - by - parts. Let $v=\sin u$, $dw = e^{u}du$, then $dv=\cos udu$, $w = e^{u}$.
By the integration - by - parts formula $\int vdw=vw-\int wdv$, we have $\int\sin(u)e^{u}du=e^{u}\sin u-\int e^{u}\cos udu$.
For $\int e^{u}\cos udu$, use integration - by - parts again. Let $v = \cos u$, $dw=e^{u}du$, then $dv=-\sin udu$, $w = e^{u}$. So $\int e^{u}\cos udu=e^{u}\cos u+\int e^{u}\sin udu$.
Let $I=\int\sin(u)e^{u}du$, then $I = e^{u}\sin u-(e^{u}\cos u + I)$.
Solving for $I$ gives $I=\frac{1}{2}e^{u}(\sin u-\cos u)+C_1=\frac{1}{2}x(\sin(\ln x)-\cos(\ln x))+C_1$ for $x\geq1$.

Case 2: When $u < 0$

The integral is $\int-\sin(u)e^{u}du$. Let $I_1=\int-\sin(u)e^{u}du$. Using integration - by - parts as above. Let $v =-\sin u$, $dw=e^{u}du$, then $dv=-\cos udu$, $w = e^{u}$.
$I_1=-e^{u}\sin u+\int e^{u}\cos udu$. For $\int e^{u}\cos udu$ using integration - by - parts again, and solving for $I_1$ gives $I_1=-\frac{1}{2}e^{u}(\sin u+\cos u)+C_2=-\frac{1}{2}x(\sin(\ln x)+\cos(\ln x))+C_2$ for $0 < x<1$.
Combining the two cases, we can write the general antiderivative as:

$$ \int\sin|\ln x|dx=
LATEXBLOCK0
$$

Answer:

$$ \int\sin|\ln x|dx=
LATEXBLOCK0
$$