The reason I'm asking this is that because $d/dx(\arccos(x))=-\frac{1}{\sqrt{1-x^2}}$, why to textbooks opt to write $\int -\frac{1}{\sqrt{1-x^2}} dx=-\arcsin(x) $ instead?
Why isn’t $\int \frac{-dx}{\sqrt{1-x^2}}$ equal to $\arccos(x)$
calculusintegration
Related Solutions
There is a basic approach here, called "trigonometric substitution." This is something to use whenever you have an integral that contains a radical of any of the following forms: $$\sqrt{a^2-x^2},\qquad \sqrt{a^2+x^2},\qquad \sqrt{x^2-a^2},\qquad a\gt 0.$$ The way trigonometric substitution works is by using the trigonometric identities $$\sin^2\theta + \cos^2\theta = 1,\qquad \tan^2\theta + 1 = \sec^2\theta.$$ The idea is to replace $x$ with something that will turn the stuff inside the square root into a square, so that you can then eliminate the radical.
Here, with $\sqrt{25-x^2}$, the idea is to use the identity $1-\sin^2\theta=\cos^2\theta$. If we replace $x$ with something that will turn $25-x^2$ into, say, $25(1-\sin^2\theta)$, then when we take the square root things will simplify.
So, set $x=5\sin\theta$. Why $5$? Because we have $\sqrt{5^2 - x^2}$.
Notice also that $-5\leq x\leq 5$ must hold for the radical to make sense. That's good, because $5\sin\theta$ can only take values there. To make this substitution reversible, we must make the assignment one-to-one (a single value of $\theta$ for each value of $x$); the easiest way to do this is to restrict ourselves to $\theta$ in the range of the $\arcsin$. So to be complete, our substitution will be $$ x = 5\sin\theta,\qquad -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}.$$ Now, what happens if we do this? Well, for the radical, we have: $$\sqrt{25 - x^2} = \sqrt{25 - 25\sin^2\theta} = \sqrt{25(1-\sin^2\theta)} = \sqrt{25\cos^2\theta} = \sqrt{(5\cos\theta)^2} = |5\cos\theta|.$$ (Don't forget the absolute values! In general, $\sqrt{a^2}=|a|$). But... because $\theta$ is supposed to be between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, then $\cos\theta\geq 0$. So $|5\cos\theta| = 5\cos\theta$, without the absolute value. So this is great! The radical completely simplifies to $$\sqrt{25-x^2} = 5\cos\theta.$$ The other bit we need to compute is $dx$, since we are doing a change of variable. Since $x=5\sin\theta$, then $dx = 5\cos\theta\,d\theta$. So in summary, we have: $$\int\sqrt{25-x^2}\,dx = \int (5\cos\theta)(5\cos\theta)\,d\theta = \int 25\cos^2\theta\,d\theta.$$ So now we need to do the integral of $\cos^2\theta$. This is a standard integral (or can be done with Weierstrass $t$-substutition to get a rational function). It can be done by using integration by parts and then solving: taking $u=\cos\theta$, $dv=\cos\theta\,d\theta$, we get: $$\begin{align*} \int\cos^2\theta\,d\theta &= \cos\theta\sin\theta +\int\sin^2\theta\,d\theta\\ &= \cos\theta\sin\theta + \int(1-\cos^2\theta)\,d\theta\\ &= \cos\theta\sin\theta + \theta - \int\cos^2\theta\,d\theta. \end{align*}$$ Solving for $\int\cos^2\theta\,d\theta$, we get $$\int\cos^2\theta\,d\theta = \frac{1}{2}\cos\theta\sin\theta + \frac{1}{2}\theta + C.$$ Going back to the original integral, we have: $$\begin{align*} \int\sqrt{25-x^2}\,dx &= \int 25\cos^2\theta\,d\theta\\ &= 25\left(\frac{1}{2}\cos\theta\sin\theta + \frac{1}{2}\theta\right) + C\\ &= \frac{25}{2}\cos\theta\sin\theta + \frac{25}{2}\theta + C. \end{align*}$$ Finally, we need to "go back to $x$". Since $x=5\sin\theta$, then $\sin\theta = \frac{1}{5}x$, $\cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1 - \frac{x^2}{25}} = \frac{1}{5}\sqrt{25-x^2}$; and $\theta = \arcsin\left(\frac{x}{5}\right)$. So we have: $$\begin{align*} \int\sqrt{25-x^2}\,dx &= \frac{25}{2}\cos\theta\sin\theta + \frac{25}{2}\theta + C\\ &= \frac{25}{2}\left(\frac{1}{5}\sqrt{25-x^2}\right)\left(\frac{x}{5}\right) + \frac{25}{2}\arcsin\left(\frac{x}{5}\right) + C\\ &= \frac{1}{2}x\sqrt{25-x^2} + \frac{25}{2}\arcsin\frac{x}{5} + C. \end{align*}$$ You can verify this by differentiation: $$\begin{align*} &\frac{d}{dx}\left(\frac{1}{2}x\sqrt{25-x^2} + \frac{25}{2}\arcsin\frac{x}{5}+C\right)\\ &=\frac{1}{2}\sqrt{25-x^2}-\frac{1}{2}\frac{x^2}{\sqrt{25-x^2}} + \frac{25}{2}\left(\frac{1}{\sqrt{ 1 - \left(\frac{x}{5}\right)^2}}\right)\frac{1}{5}\\ &= \frac{1}{2}\sqrt{25-x^2} - \frac{1}{2}\frac{x^2}{\sqrt{25-x^2}} + \frac{5}{2}\left(\frac{1}{\frac{1}{5}\sqrt{25-x^2}}\right)\\ &= \frac{1}{2}\sqrt{25-x^2} - \frac{1}{2}\frac{x^2}{\sqrt{25-x^2}} + \frac{25}{2}\frac{1}{\sqrt{25-x^2}}\\ &= \frac{1}{2}\sqrt{25-x^2} + \frac{1}{2}\left(\frac{25 - x^2}{\sqrt{25-x^2}}\right)\\ &= \frac{1}{2}\sqrt{25-x^2} + \frac{1}{2}\sqrt{25-x^2} = \sqrt{25-x^2}. \end{align*}$$
In general:
If you have a radical $\sqrt{a^2-x^2}$, you can try a substitution $x = a\sin\theta$, $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}$, which will transform $\sqrt{a^2-x^2}$ into $a\cos\theta$.
If you have a radical $\sqrt{a^2+x^2}$, you can try a substitution $x=a\tan\theta$, with $-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}$, which will transform $\sqrt{a^2+x^2}$ into $a\sec\theta$.
If you have a radical $\sqrt{x^2-a^2}$, you can try a substitution $x = a\sec\theta$ with $-\frac{\pi}{2}\lt x \lt \frac{\pi}{2}$, which will transform $\sqrt{x^2-a^2}$ into $a\tan\theta$.
Once you do the substitution, you'll have an integral involving sines and cosines; these can be solved, in the worst case scenario, by using Weierstrass's $t$-substitution to change them into integrals of rational functions and doing partial fractions; and sometimes just directly.
There are other alternatives: you can use hyperbolic functions, for example, which satisfy $$\cosh^2\theta - \sinh^2\theta = 1,\qquad 1 - \tanh^2\theta = \frac{1}{\cosh^2\theta},$$ so you can use "hyperbolic substitution" instead of trigonometric substitution. The result may need further transformation (just as in the trigonometric case).
There are a couple of ways to see this. Firstly, draw a right triangle, call it $ABC$ (with $C$ being the right angle), with side lengths $a$, $b$ and $c$ with the usual convention. Then $\arcsin(\frac{b}{c})$ is the measure of the angle $CBA$. Additionally, $\arccos(\frac{b}{c})$ is the angle of the angle of the opposite angle $CAB$, so $\arccos(\frac{b}{c}) = \frac{\pi}{2}-\arcsin(\frac{b}{c})$ since the opposite angles must sum to $\frac{\pi}{2}$. From here, you get the result.
We could also do some calculus to figure it out. Let's let $f(x) = \arcsin(x)+\arccos(x)$. Then $f'(x) = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0$. Thus $f$ is constant. What is $f(0)$ equal to?
Best Answer
They are both correct, since the two answers differ by a constant. This is because $$\arccos(x)+\arcsin(x)=\pi/2.$$