show that : $\int \sqrt{a^2-x^2}dx = \frac{1}{2}x\sqrt{a^2-x^2} + \frac{1}{2}a^2 \arcsin(\frac{x}{a}) + C \ (a > 0)$

Question

How to show that : $\int \sqrt{a^2-x^2}dx = \frac{1}{2}x\sqrt{a^2-x^2} + \frac{1}{2}a^2 \arcsin(\frac{x}{a}) + C \ (a > 0)$

My first idea is to take the derivative of both sides. The right side will yield $\sqrt{a^2-x^2}$, the left side is a bit harder to deal with. If $a > 0$ and $a > x$ in order to avoid a complex result, then I can safely say that $x > 0$. Hence we can create a new function for the integral:

$$F(x) = \int_{0}^{x} \sqrt{a^2-x^2}dx$$

By the fundamental theorem of calculus, the derivative of this function is $\sqrt{a^2-x^2}$. Hence showing that the two results are equal on both sides.

Is this solution correct?

Answer 1

Here it is another way to solve it. Let $x = a\sin(\theta)$. Thus we have \begin{align*} \int\sqrt{a^{2} - x^{2}}\mathrm{d}x & = a\int\sqrt{a^{2} - a^{2}\sin^{2}(\theta)}\cos(\theta)\mathrm{d}\theta\\\\ & = a^{2}\int\sqrt{1-\sin^{2}(\theta)}\cos(\theta)\mathrm{d}\theta\\\\ & = a^{2}\int\cos^{2}(\theta)\mathrm{d}\theta\\\\ & = a^{2}\int\frac{\cos(2\theta) +1}{2}\mathrm{d}\theta\\\\ & = a^{2}\left[\frac{\sin(2\theta)}{4} + \frac{\theta}{2}\right] + c \end{align*}

Given that $x = a\sin(\theta)$, this implies that \begin{align*} \begin{cases} \displaystyle\theta = \arcsin\left(\frac{x}{a}\right)\\\\ \displaystyle\sin(2\theta) = 2\sin(\theta)\cos(\theta) = \frac{2x\sqrt{a^{2} - x^{2}}}{a^{2}} \end{cases} \end{align*} and we are done.

Hopefully this helps!

Answer 2

Integrate by parts as follows \begin{align*} \int\sqrt{a^{2} - x^{2}}\mathrm{d}x & = \int\frac{\sqrt{a^{2} - x^2}}{2x}d(x^2) \\ & = \frac{1}{2}x\sqrt{a^2-x^2} +\frac12 a^2\int\frac{1}{\sqrt{a^{2} - x^2}}dx\\ & = \frac{1}{2}x\sqrt{a^2-x^2} + \frac{1}{2}a^2 \arcsin\frac{x}{a}+C\\ \end{align*}