Added: I interpreted the question as follows: suppose that we use the integration by parts formula,
$$\int u(x)v'(x)\,dx = u(x)v(x) - \int v(x)u'(x)\,dx$$
and $\int v(x)u'(x)\,dx = \alpha \int u(x)v'(x)\,dx$ for some constant $\alpha$, so that we can "solve" for the original integral as
$$\int u(x)v'(x)\,dx = \frac{1}{1+\alpha}u(x)v(x) + C.$$
Is it the case that we can solve the original integral by doing a direct substitution instead of integration by parts?
I think this works:
Suppose that $v(x)u'(x) = \alpha u(x)v'(x)$ for some $\alpha\neq -1$. (I think this is essentially what you have, since the integral on the left hand side is $\int u(x)v'(x)\,dx$, and the integral on the right is $\int v(x)u'(x)\,dx$). I restrict to $\alpha\neq -1$, because if $\alpha=-1$, then you cannot "solve" for the original integral. But in fact, from the work below we will get that $u=\frac{B}{|v|}$, so that integration by parts will simply result in the true (but useless) $\int u\,dv = B + \int u\,dv$.
If $u$ or $v$ are zero, then the original integral was the integral of $0$, so we may discard that case. So we get that $\frac{u'}{u} = \alpha \frac{v'}{v}$, and integrating we get $\ln|u|=\alpha\ln|v|+C$, or $|u| = A|v|^{\alpha}$ for some $A\gt 0$; hence $u=B|v|^{\alpha}$ for some $B\neq 0$. So the original integral was
$$\int u(x)v'(x)\,dx = \int B|v(x)|^{\alpha}v'(x)\,dx$$
which suggests the substitution $g=v(x)$ to get $B\int |g|^{\alpha}\,dg$.
Added: Note that if $\alpha = -1$, we still get $u=B|v|^{-1}$, and
so again the substitution $g=|v(x)|$ will do the trick.
You have an algebra error here:
$$\begin{align*}
\int t^3 e^{-t^2}dt&=\int u^{3/2} e^{-u} \frac{du}{2u^{1/2}}\\
&=\int u^2 e^{-u}\frac{du}2\;:
\end{align*}$$
$\dfrac{u^{3/2}}{u^{1/2}}=u$, not $u^2$.
Best Answer
Hint: Let $u = \ln{x}$. Then $du = \frac{1}{x} dx$, or $x du = dx$. Using the fact that $x = e^u$, we can write
$$\int \sin \ln{x} dx = \int e^u \sin{u}du$$
This is a common integral that can be done by using parts twice, or by recognizing that
$$\sin{u} = \operatorname{Im} e^{iu}$$