integral with several answers

Question

Could you help me please with a question about integrals? Can an integral have more than one answer? For example with this integral: $$\int\sqrt{1+\sqrt{1-x^2}}dx$$ Doing by replacing u=$\sqrt{1-x^2}$, I have this solution: $$2\sqrt{1-\sqrt{1-x^2}}-\frac23 \left(1-\sqrt{1-x^2} \right)^{3/2},$$ and another solution doing by replacing $\sin \theta =x/1$, I get: $$2\sqrt{2} \sin(\arcsin(x/2))-\frac43 \sqrt{2} \sin^3(\arcsin(x/2))).$$ The graph of each one is in this picture: Thanks in advance.

Answer 1

The error in one of your solutions lies in the fact that you have to consider the sign of the $x$ values used when performing the integration. In particular, using $u=\sqrt{1-x^2}$, we cannot say that $x=\sqrt{1-u^2}$ because $x$ may be negative whereas the square root term cannot. Also I believe you forgot to use the fact that $\sqrt{x^2}=|x|$ in your working. The trigonometric substitution has given you the correct answer for the integral in the interval $x\in[-1,1]$. We can get the correct algebraic solution by using the identity $$\sin{\left(\frac{\arcsin{(x)}}2\right)}=\text{sgn}(x)\sqrt{\frac{1-\sqrt{1-x^2}}2}$$ for $x\in[-1,1]$. Here $\text{sgn}(x)$ denotes the Sign function. This gives the value of the integral as $$2\text{sgn}(x)\sqrt{1-\sqrt{1-x^2}}-\frac23\text{sgn}(x) (1-\sqrt{1-x^2})^{3/2}+C$$ instead of your solution. The two answers are then the exact same function.

Answer 2

Yes, an integral can be have more than one "answer", hence the importance of the constant that is added when finished integrating. More correctly, if two functions $f$ and $g$ they have the same derivative, so we can say that these differ by a constant : $$f'=g' \, \Rightarrow \, f=g+c \, \textrm{ for some } c$$ Think of a simpler example, let $$\int 2 \sin x \cos x \ dx$$ then, if we take $u=\sin x$, this integral is equal to $\sin^2 x$ and if we take $u=\cos x$ the integral is $-\cos^2 x$.

As you can see, the integral has two (or more) answers, and that clearly $\sin^2 x \neq -\cos^2 x$.

In fact, this two functions differ by a constant, by $1$ : $$\sin^2 x = -\cos^2 x \color{red}{+1}$$