2
$\begingroup$

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

$\endgroup$
5
  • 2
    $\begingroup$ You're forgetting $+C$ which is important here ;) $\endgroup$
    – user223391
    May 17, 2016 at 16:49
  • 1
    $\begingroup$ Note that $\arcsin t+\arccos t=\frac{\pi}{2}$ $\endgroup$
    – egreg
    May 17, 2016 at 16:50
  • $\begingroup$ +C, and once you account for that you will see that the to answers are the same. $\endgroup$
    – Doug M
    May 17, 2016 at 16:50
  • $\begingroup$ 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. $\endgroup$
    – MrYouMath
    May 17, 2016 at 16:53
  • $\begingroup$ Thanks. Now I realize that the constant is very important here. I hardly ever be aware of a constant $C$ in the indefinite integral... $\endgroup$
    – IgNite
    May 17, 2016 at 16:55

1 Answer 1

6
$\begingroup$

$\cos(\pi/2-x)=\sin(x)$ 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.

$\endgroup$
4
  • $\begingroup$ Thanks for your answer. Oh I'm sorry, I shouldn't have asked this question. Anyone knows how to delete this post? $\endgroup$
    – IgNite
    May 17, 2016 at 16:53
  • $\begingroup$ If you're allowed to, it is directly below the question, there would be buttons "share cite edit delete flag". I'm not sure you can at your rep level, though. $\endgroup$
    – Ian
    May 17, 2016 at 16:54
  • 1
    $\begingroup$ @ignite why shouldn't you have asked the question? Don't delete it. $\endgroup$
    – user223391
    May 17, 2016 at 16:55
  • $\begingroup$ I think this is some common pitfall in calculations of the indefinite integrals, but hope that this will be helpful for someone who might probably encounter the same problem. (Now I need more 17 seconds to accept the answer :)) $\endgroup$
    – IgNite
    May 17, 2016 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.