# Integration by parts resulting in answer equal to the question itself

$$\int\frac{x\arcsin x}{\sqrt{1 - x^2}}dx = I$$

By applying integration by parts several times on this question i get the answer to be equal to the question itself i.e. $$0 = 0$$. I will try to demonstrate how i got stuck in the strange problem.

$$x(\arcsin x)(\arcsin x) - \int(\arcsin x)(\arcsin x+\frac{x}{\sqrt{1 - x^2}})dx = I$$

$$x(\arcsin x)^2 - \int(\arcsin x)^2- \int \frac{x\arcsin x}{\sqrt{1 - x^2}} dx = I$$

as $$I=\int\frac{x\arcsin x}{\sqrt{1 - x^2}}dx$$ we get

$$x(\arcsin x)^2 - \int(\arcsin x)^2 = 2I$$

Now again i will apply integration by parts on $$\int(\arcsin x)^2dx$$

$$x(\arcsin x)^2 - [x(\arcsin x)^2- 2\int \frac{x\arcsin x}{\sqrt{1 - x^2}} dx] = 2I$$

$$x(\arcsin x)^2 - x(\arcsin x)^2 + 2\int\frac{x\arcsin x}{\sqrt{1 - x^2}} dx = 2I$$

As $$\int\frac{x\arcsin x}{\sqrt{1 - x^2}}dx = I$$ so we get

$$2\int\frac{x\arcsin x}{\sqrt{1 - x^2}}dx = 2I$$

Or The final solution is :

$$2I = 2I$$ ?????

I never have heard or seen this thing before and can find it nowhere on the internet or in any book. Is anyone so mathematically inclined to explain this stuff.

NOTE : I am not asking you to solve this question, as I have already solved this using some other technique which resulted in the correct answer, i.e.,

$$x - \arcsin x{\sqrt{1 - x^2}} + c$$

I just need an answer to the last statement ( $$2I = 2I$$ ) that just occured using the above procedure. I am a student of computer science and am in my early years of education.Please try to explain in a way that would best suit my experience (Something that i could easily understand) as i don't know advanced mathematics.

• Integrating by parts is moving derivatives from one function to another. What you do is that you first integrate by parts, and then do it back again. $$\int f' g=fg-\int fg'.$$ It is not so strange that if you start with $I$, integrate by parts, and then integrate by parts "back" again, that you end up with the same thing as you started with. Commented Apr 5, 2016 at 14:59
• To actually do this one using integration by parts, take $u=\arcsin x$ and $dv = x\,dx/\sqrt{1-x^2}$. Commented Apr 5, 2016 at 15:23
• @Gedgar as i have already said that i have solved this question.Simply i want the answer to the unusual occurance in the end.This is the very first time i have encountered such a thing.This question might be solved using many different techniques all resulting in the same answer, but why the procedure adopted above results in The question being equal to itself. Commented Apr 6, 2016 at 2:52
• @mickep i know you can get back the same question while reapplying integration by parts many times but that helps in the final solution here it just results in an obvious thing ("Ofcourse the question is equal to the question or 0 = 0"). Commented Apr 6, 2016 at 3:48

Try and solve the integral $$\int_0^1 e^{2x} dx$$ by first substituting $$u = 2x$$ to obtain $$\int_0^2 2 e^{u} du$$, and then (foolishly) substituting $$v = \frac{1}{2} u$$ to obtain $$\int_0^1 e^{2v} dv$$. Voilà, out pops the original question.
Applying $$u\,v = \int {d(uv)} = \int {u\,dv} + \int {v\,du}$$ two times may well result in a "circle" $0=0$ as with
$I=a+b=a+(I-a)$ => $I=I$
• When integrating by parts, you have (at least) two choices for $u$ and $dv$. If the second integration by parts gets you back where you started, it means you made the wrong choice the second time (so it undid the first time). You may be able to continue by switching your choice for $u$ and $dv$ for the second integration by parts. Commented Apr 5, 2016 at 15:18