Question

(int sin xln|x|dx)

Explanation:

Step1: Use integration by parts

Let $u = \ln x$, $dv=\sin xdx$. Then $du=\frac{1}{x}dx$, $v = -\cos x$.

Step2: Apply integration - by - parts formula $\int u dv=uv-\int v du$

$\int\ln x\sin xdx=-\ln x\cos x+\int\frac{\cos x}{x}dx$.
The integral $\int\frac{\cos x}{x}dx$ is the cosine integral $\text{Ci}(x)$. So the integral $\int\ln x\sin xdx=-\ln x\cos x+\text{Ci}(x)+C$.

Answer:

$-\ln x\cos x+\text{Ci}(x)+C$