# Integral of $x/\sqrt{a^2 - x^2}$ [closed]

I have been attempting to solve for the integral of the below equation but have had no luck so far. After searching online I have only managed to find one mention of it but with no explanation (the below picture).

I have tried using the quotient rule by parts formula, and substitution but have found myself running circles each time and getting nowhere near the proposed answer on the site. I need this integral for a larger integral I'm doing but need to know its proof so I may properly explain its use in my paper.

I know I'm not really providing much here, but I've become completely lost with this integral. Any help is much appreciated.

• Hint: $t=x^2+a^2$. Sep 11, 2015 at 1:42
• Next time, just state the problem in short Sep 11, 2015 at 1:51
• The formula in the title subtracts $x^2$ from $a^2$, but the formula in the picture in the question body adds $x^2$ to $a^2$. Which integral do you want? Sep 11, 2015 at 2:05
• The proof of the equation in the picture is simply the fact that the derivative of the expression on the right-hand side of the equation is the expression inside the integral on the left-hand side. What do you need other than that to "explain its use" in your paper? Sep 11, 2015 at 2:28
• Welcome to Maths.SE! Please do not use pictures for critical portions of your post. Pictures cannot be searched and are inaccessible to those using screen readers. Sep 13, 2015 at 19:43

Let $u = x^2+a^2$ as has been suggested. Then $\frac{du}{dx} = 2x$ or equivalently $dx = \frac{du}{2x}$. Substituting these quantities into the integral gives

\begin{align} \int \frac{1}{2\sqrt{u}} du = \sqrt{u} + C = \sqrt{x^2+a^2} + C \end{align}

• Okay this fixes a lot, because I tried using t = a^2 +x^2 but I messed up on the substitution. Silly mistake, but thanks a lot, you and Lucian help a lot Ill accept answer in a sec after i look this over a bit and properly understand what I did wrong. Sep 11, 2015 at 2:19
• @23scurtu No problem, let me know if you don't understand something. Sep 11, 2015 at 2:21

You need only the "fundamental theorem of calculus" here!

Note that $$\int_{x} \frac{x}{\sqrt{x^{2}+a^{2}}} = \int_{x}D(x^{2}+a^{2})^{1/2} = (x^{2}+a^{2})^{1/2} + \text{some constant}.$$

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 11, 2015 at 7:01

Let's use U-substitution to solve this problem!

$$\int \frac{x}{\sqrt{x^{2}+a^{2}}}dx$$

$$\text{Let: }\qquad u = x^2+a^2\quad \text{and}\quad du = 2xdx$$

Therefore, \begin{align} \require{cancel}\\ \int \frac{x}{\sqrt{x^{2}+a^{2}}}dx &= \frac{1}{2}\int\frac{1}{\sqrt u}du\\ &= \frac{1}{\cancel2} (\cancel2\sqrt u) + C\\ &= \sqrt u + C\\ &= \sqrt{x^2+a^2} + C\\ \end{align}

• when you canceled out the 2, I thought that there was only one 2 that came from the 1/2. where did the second one come from? Sep 11, 2015 at 2:32
• The $\int \frac{1}{\sqrt u}du$ is $2\sqrt u + C$. Do you want me to explicitly have that worked out in the answer? Sep 11, 2015 at 2:35
• Oh crap! I see how it works. raising the power to the u^-1/2 requires multiplying by 2 to cancel out the 1/2 that would be carried down through the derivation of u^1/2. Adding that would prob help anyone else coming by this question. Sep 11, 2015 at 2:46
• @23scurtu Don't think of it as multiplying by 2 to cancel out the $\frac{1}{2}$ in the long run, just think of the short term integral, $\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=\frac{u^{\frac{1}{2}}}{\frac{1}{2}}=2u^{\frac{1}{2}} = 2\sqrt u$. Sep 11, 2015 at 2:52
• alright thankks Sep 11, 2015 at 3:07