calculus - graphical meaning of integration by parts of this function

I have seen this and this. After looking at the first link I was pretty satisfied with the answer. Now I understand why

$\int vdu=uv-\int udv$

works graphically.

I have even worked out $\int (\sin^{-1}x)dx$ using integration by parts and understood it graphically.

But I am not able to visualize this $\int (x\sin x)dx$ in the graphical manner as I'm not able to find an inverse which is important for the graphical method to work. However, I could solve it mathematically.

Now my questions are

i) Is the graphical intuition only limited to functions which can be integrated this way?

$\int f^{-1}(y)dy=xy-\int f(x)dx$

ii)Please help me with a graphical intuition behind solving functions like $\int (x\sin x)dx$ using integration by parts (if any). I will try to do the remaining math myself.

Any suggestion is of great help to me. Thank you.

Answer

I am not sure if this helps, but I'll show you a proof for integration by parts then use it to evaluate $\int x\sin x\,dx$. I don't know how to really think of integration by parts graphically, most of the visualization for me comes from seeing the algebra.

Let $u(x)$ and $v(x)$ be differentiable, continuous functions. Then we know that
$$\frac{d}{dx}\left[u(x)v(x)\right]=u'(x)v(x)+u(x)v'(x)$$
And if we integrate both sides, we have that
$$\int \frac{d}{dx}\left[u(x)v(x)\right] dx=\int u'(x)v(x)dx+\int u(x)v'(x)dx$$
$$u(x)v(x)=\int u'(x)v(x)dx+\int u(x)v'(x)dx$$
And after some simple re-arrangement,

$$\int u(x)v'(x)dx=u(x)v(x)-\int u'(x)v(x)dx$$
Which is usually just abbreviated as
$$\int udv=uv-\int vdu$$
Which completes our proof.

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Next, we evaluate some examples.

Ex 1: $$I=\int x\sin(x)dx$$

For this, I do not recommend trying to visualize the integration, because that could just make things more complicated. Instead integrate by parts with
$$u=x\Rightarrow du=dx\\ dv=\sin(x)dx\Rightarrow v=-\cos(x)$$
Hence,
$$I=-x\cos(x)+\int\cos(x)dx$$
$$I=-x\cos(x)+\sin(x)+C$$
Which is rather easy: no visualization required.

Ex. 2:
$$I=\int x\tan^{-1}(x)dx$$
I chose this example because it is rather hard to visualize, but can be computed with integration by parts as follows:

$$u=\tan^{-1}(x)\Rightarrow du=\frac{dx}{1+x^2}\\ dv=xdx\Rightarrow v=\frac{x^2}2$$
Hence
$$I=\frac{x^2}2\tan^{-1}(x)-\frac12\int\frac{x^2}{1+x^2}dx$$
This final integral is easier than it looks, and it can be computed by noting that
$$\begin{align} \int\frac{x^2}{x^2+1}dx&=\int\frac{x^2+1-1}{x^2+1}dx\\ &=\int\frac{x^2+1}{x^2+1}dx-\int\frac{dx}{x^2+1}\\ &=\int dx-\int\frac{dx}{x^2+1}\\ &=x-\tan^{-1}(x)\\ \end{align}$$

So, without having to visualize anything,
$$I=\frac{x^2+1}2\tan^{-1}(x)-\frac{x}2+C$$
Tell me if there's anything I can do to improve my answer if it is not what you needed.