# Evaluate $\int{\sqrt{a^2 - x^2}}dx$

I'm trying to solve the following integral, but seems these 2 methods led to different answers. I think one of the methods must be incorrect. But why doesn't one of them work?

Evaluate $\int{\sqrt{a^2 - x^2}}\ dx$

My friend evaluated this way:

First let $x=a\cos{\theta}$, so $a^2-x^2=a^2(1-\cos^2\theta)=a^2\sin^2{\theta}$ $$\int{\sqrt{a^2 - x^2}}\ dx = -\int{a\sin{\theta}}\ d(a\cos{\theta}) = -\int{a^2\sin^2{\theta}}\ d\theta=-a^2\int{\frac{1-\cos{2\theta}}{2}}\ d\theta$$ $$= -\frac{a^2}{2}\int(1-\cos{2\theta})\ d\theta = -\frac{a^2}{2}\left(\theta - \frac{\sin{2\theta}}{2}\right)$$ $$= \frac{a^2}{2}\left(-\cos^{-1}{\frac{x}{a}}+\frac{{x}\sqrt{a^2-x^2}}{a^2}\right)$$

However I've done this way:

First let $x=a\sin{\theta}$, so $a^2-x^2=a^2(1-\sin^2\theta)=a^2\cos^2{\theta}$ $$\int{\sqrt{a^2 - x^2}}\ dx = \int{a\cos{\theta}}\ d(a\sin{\theta}) = \int{a^2\cos^2{\theta}}\ d\theta=a^2\int{\frac{1+\cos{2\theta}}{2}}\ d\theta$$ $$= \frac{a^2}{2}\int(1+\cos{2\theta})\ d\theta = \frac{a^2}{2}(\theta + \frac{\sin{2\theta}}{2})$$ $$= \frac{a^2}{2}\left(\sin^{-1}{\frac{x}{a}}+\frac{{x}\sqrt{a^2-x^2}}{a^2}\right)$$

• You're forgetting $+C$ which is important here ;)
– user223391
May 17, 2016 at 16:49
• Note that $\arcsin t+\arccos t=\frac{\pi}{2}$ May 17, 2016 at 16:50
• +C, and once you account for that you will see that the to answers are the same. May 17, 2016 at 16:50
• Like Ian answered, both expressions are the same. But I would rather use the first substitution as the second has to consider the magnitude of $\cos\theta$ on the interval from $0$ to $\pi$. The second one is better as you don't have to consider the magnitude of $\sin\theta$ on the same interval. May 17, 2016 at 16:53
• Thanks. Now I realize that the constant is very important here. I hardly ever be aware of a constant $C$ in the indefinite integral... May 17, 2016 at 16:55

$\cos(\pi/2-x)=\sin(x)$ and vice versa. If you haven't seen this before, the geometric explanation is that the sine of one of the acute angles in a right triangle is the cosine of the other.
Therefore $-\cos^{-1}(x)$ and $\sin^{-1}(x)$ are the same up to a constant. Since indefinite integrals are only defined up to a constant (and the factor on the outside is a constant), your two solutions are consistent with each other.