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.
Best 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.
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.