Other method other than integration by parts I would like to integrate this
$$\int x \ln x\sin x\mathrm dx$$

can I do it with another quick method than using integration by part?
Here is a fun but not very useful trick: if you know the antiderivative already, any integral can be computed using only $u$-substitution. Proof:

$$\int f(x) \, dx$$
Let $F'(x) = f(x)$.
$$u = F(x), du = f(x) \, dx$$
$$\int du = u + C = F(x) + C.$$

So for your integral, we see that it can be solved using the magic $u$-substitution

$$u = -Si(x) + \sin x + \ln x (\sin x - x \cos x).$$

We have

$$du = x \ln x \sin x \, dx$$
$$\int du = u + C = -Si(x) + \sin x + \ln x (\sin x - x \cos x),$$

as desired.
Here is a fun but not very useful trick: if you know the antiderivative already, any integral can be computed using only $u$-substitution. Proof:

$$\int f(x) \, dx$$
Let $F'(x) = f(x)$.
$$u = F(x), du = f(x) \, dx$$
$$\int du = u + C = F(x) + C.$$

So for your integral, we see that it can be solved using the magic $u$-substitution

$$u = -Si(x) + \sin x + \ln x (\sin x - x \cos x).$$

We have

$$du = x \ln x \sin x \, dx$$
$$\int du = u + C = -Si(x) + \sin x + \ln x (\sin x - x \cos x),$$

as desired.