26. $intsin(ln x)dx$

Explanation:

Step1: Use substitution

Let $u = \ln x$, then $x=e^{u}$ and $dx = e^{u}du$. The integral becomes $\int\sin(u)e^{u}du$.

Step2: Apply integration - by - parts

The formula for integration by parts is $\int f\mathrm{d}g=fg-\int g\mathrm{d}f$. Let $f = \sin(u)$ and $\mathrm{d}g=e^{u}du$. Then $\mathrm{d}f=\cos(u)du$ and $g = e^{u}$. So $\int\sin(u)e^{u}du=\sin(u)e^{u}-\int\cos(u)e^{u}du$.

Step3: Apply integration - by - parts again

For $\int\cos(u)e^{u}du$, let $f=\cos(u)$ and $\mathrm{d}g = e^{u}du$. Then $\mathrm{d}f=-\sin(u)du$ and $g = e^{u}$. So $\int\cos(u)e^{u}du=\cos(u)e^{u}+\int\sin(u)e^{u}du$.

Step4: Solve for $\int\sin(u)e^{u}du$

We have $\int\sin(u)e^{u}du=\sin(u)e^{u}-\left(\cos(u)e^{u}+\int\sin(u)e^{u}du\right)$. $2\int\sin(u)e^{u}du=\sin(u)e^{u}-\cos(u)e^{u}+C$. $\int\sin(u)e^{u}du=\frac{1}{2}e^{u}(\sin(u)-\cos(u))+C$.

Step5: Substitute back $u = \ln x$

Since $u=\ln x$ and $e^{u}=x$, the original integral $\int\sin(\ln x)dx=\frac{1}{2}x(\sin(\ln x)-\cos(\ln x))+C$.

Answer:

$\frac{1}{2}x(\sin(\ln x)-\cos(\ln x))+C$