I'll restate the accepted answer in different notation, which is easier for me to parse: let
$$u=\sqrt{1-x^2}, \quad dv=dx$$
so that
$$du=\frac{-x}{\sqrt{1-x^2}}dx,\quad v=x$$
For brevity, write $I=\int \sqrt{1-x^2}\, dx$. Using $\int u\,dv = uv-\int v\,du$, obtain
$$
I = x\sqrt{1-x^2} - \int \frac{-x^2}{\sqrt{1-x^2}} \,dx
$$
The last integral does not look simpler than $I$ itself, but it can be related back to it:
$$
\int \frac{-x^2}{\sqrt{1-x^2}} \,dx = \int \frac{1-x^2}{\sqrt{1-x^2}} \,dx - \int \frac{1 }{\sqrt{1-x^2}} \,dx = I - \sin^{-1}x
$$
So,
$$I = x\sqrt{1-x^2} - (I-\sin^{-1}x)$$
and solving for $I$ yields
$$\int \sqrt{1-x^2}\, dx = \frac12 x\sqrt{1-x^2} + \frac12 \sin^{-1}x + C$$
For completeness and comparison, I'll add the conventional solution using $x=\sin t$ substitution. Here $dx=\cos t\,dt$, so
$$
\int\sqrt{1-x^2}\,dx = \int \cos^2 t\,dt =\int \left(\frac12+\frac{\cos 2t}{2}\right)\,dt = \frac{t}{2}+\frac{\sin 2t}{4}+C $$
To return to $x$, note that $t=\sin^{-1}x$ and $\sin 2t = 2\sin t\cos t = 2x\sqrt{1-x^2}$.
Based on your attempt, we can integrate by parts the proposed function to obtain:
\begin{align*}
\int x\exp(\sin(x))\cos(x)\mathrm{d}x = x\exp(\sin(x)) - \int\exp(\sin(x))\mathrm{d}x
\end{align*}
Unfortunately, the last integral cannot be solved in terms of elementary functions.
Hopefully this helps!
Best Answer
In order to prove $\int \log x \, \mathrm{d}x = x \log x - x + \mathsf{C}$, it suffices to show that
$$ \int_{1}^{x} \log t \, \mathrm{d}t = x \log x - x + 1. $$
We establish this with different methods, avoiding integration by parts technique and 'guessing the antiderivative' strategy.
Method 1. Assume $a \geq 1$. Then by Fubini's theorem,
\begin{align*} \int_{1}^{a} \log x \, \mathrm{d}x &= \int_{1}^{a} \int_{1}^{x} \frac{1}{t} \, \mathrm{d}t\mathrm{d}x \\ &= \int_{1}^{a} \int_{t}^{a} \frac{1}{t} \, \mathrm{d}x\mathrm{d}t \\ &= \int_{1}^{a} \left( \frac{a}{t} - 1 \right) \, \mathrm{d}t \\ &= a \log a - a + 1 \end{align*}
Similar computation shows that the above result continues to hold for $0 < a < 1$.
(In reality, however, this computation still bears the flavor of integration-by-parts technique.)
Method 2. (Regularizing) It can be shown that $(x^{\epsilon} - 1)/\epsilon$ converges uniformly to $\log x$ as $\epsilon \to 0^+$ on any compact subset of $(0, \infty)$. Then
$$ \int_{1}^{x} \log t \, \mathrm{d}t = \lim_{\epsilon \to 0^+} \int_{1}^{x} \frac{t^{\epsilon} - 1}{\epsilon} \, \mathrm{d}t = \lim_{\epsilon \to 0^+} \left( \frac{x^{\epsilon+1}-1}{\epsilon(\epsilon+1)} - \frac{x-1}{\epsilon} \right) = x\log x - x + 1. $$
Method 3. (Solving functional equation) Define the function $f : (0, \infty) \to \mathbb{R}$ by $f(x) = \int_{1}^{x} \log t \, \mathrm{d}t$. Then for $a, b > 0$,
\begin{align*} f(ab) &= \int_{1}^{a} \log t \, \mathrm{d}t + \int_{a}^{ab} \log t \, \mathrm{d}t \\ &= \int_{1}^{a} \log t \, \mathrm{d}t + \int_{1}^{b} a \log (as) \, \mathrm{d}s \tag{$t = as$}\\ &= f(a) + a(b-1) \log a + a f(b). \end{align*}
By switching the role of $a$ and $b$, we also get $f(ab) = f(b) + b(a-1)\log b + bf(a)$, and so,
$$ f(a) + a(b-1)\log a + af(b) = f(b) + b(a-1)\log b + b f(a). $$
Rearranging the identity, for $a, b \neq 1$ we get
$$ \frac{f(a) - a\log a}{a-1} = \frac{f(b) - b\log b}{b-1}. $$
The right-hand side can be further simplified, using $f(1) = 0$ あand $\log 1 = 0$, as
$$ = \frac{f(b) - f(1)}{b-1} - \frac{\log b - \log 1}{b-1} - \log b. $$
Together with $f'(1) = \log 1 = 0$ and $(\log x)'|_{x=1} = 1$, this converges to $-1$ as $b \to 1$. So it follows that
$$ f(a) = a \log a - (a - 1). $$