8
$\begingroup$

$$I =\int\frac{1}{\sqrt{25-x^2}}dx$$ enter image description here

($\theta$ on left corner) $$\tag 1 5\cos(\theta)=x$$ $$\tag 2 5\sin(\theta)=\sqrt{25-x^2}$$ $$\tag 1 -5\sin(\theta)\,d\theta=dx$$ $$I=\int\frac{1}{5\sin(\theta)} \cdot (-5) \sin(\theta) \, d\theta$$ $$I=\int-d\theta$$ $$\tag 1 \theta=\arccos(x/5)$$ $$I = -\arccos\left(\frac{x}{5}\right)+c$$ However, putting this integral into WA gives $\arcsin\left(\frac{x}{5}\right)+c$ and two are clearly not equivalent.

$\endgroup$
  • 1
    $\begingroup$ "and two are clearly not equivalent". Sure they are. Sin and cos functions are the same but for starting point shift. Consider for $\cos(\frac \pi 2 - x)= \sin x$. So $\arcsin x = -arccos x + \frac \pi 2$ $\endgroup$ – fleablood Feb 16 at 16:18
  • $\begingroup$ Actually just look at a graph of $\arcsin$ and $\arccos$. It's very clear that $\arcsin x= -\arccos x+\frac \pi 2$ $\endgroup$ – fleablood Feb 16 at 16:33
  • 2
    $\begingroup$ How in the world did you conclude that the two are not equivalent? A trigonometric identity says that if $0\le x\le1$ then $$ \arcsin x + \arccos x = \frac \pi 2. $$ $\endgroup$ – Michael Hardy Feb 16 at 17:14
5
$\begingroup$

It is actually the same, since in wolframalpha instead of substituting $x=5\cos\theta$ it used $x=5\sin\theta$, you can try or observe that the substitution gives you the same result

$\endgroup$
  • 3
    $\begingroup$ Note proper notation: $$ \text{right:} \quad 5\cos\theta $$ $$ \text{wrong:} \quad 5cos\theta $$ Writing the code as \cos (similarly \sin, \log, \exp, \max, \det, etc.) does not only prevent italicization, but results in proper spacing. And the spacing depends on the context, thus in the first line below you see more space to the right of $\cos$ than in the second: $$ \begin{align} & 5\cos\theta \\ & 5\cos(\theta) \end{align} $$ $\endgroup$ – Michael Hardy Feb 16 at 17:17
12
$\begingroup$

Your answer and WA's are equivalent.

$\arccos \dfrac x5 + \arcsin \dfrac x5 = \dfrac {\pi}2$, so $\arcsin \dfrac x5 = \dfrac {\pi} 2 - \arccos \dfrac x5$,

so $\arcsin \dfrac x5 + C_1 = -\arccos \dfrac x5 + C_2$, where $C_1-C_2=\dfrac\pi2$.

$\endgroup$
6
$\begingroup$

minus arccosine and sine of x for x in -5 to 5 It appears that $-\arccos \frac{x}{5} + C$ and $\arcsin \frac{x}{5} + C$ are the same collections of functions. (Recall that an antiderivative is an infinite collection of functions differing only in vertical translation. The "${}+C$" is to remind us that we have obtained a collection of functions.)

The plotted result is a consequence of the trigonometric reflection identity $\sin(\pi/2 - \theta) = \cos \theta$ in the form $$ \sin(\theta - \pi/2) = -\cos \theta $$so we see arcsine shifted $\pi/2$ units above minus arccosine.

$\endgroup$
  • $\begingroup$ Nice picture; +1 $\endgroup$ – J. W. Tanner Feb 17 at 2:52
4
$\begingroup$

Note: $\cos(\frac \pi 2-x) = \sin x$

So if $\sin x = M$ whe $-\frac \pi 2 < x < \frac \pi 2$ then $x = \arcsin M$.

And if $\sin x = \cos(\frac \pi 2-x)=M$ then $0 < \frac \pi 2- x < \pi$ and $\frac \pi 2- x =\arccos M$ and $x = -\arccos M -\frac \pi 2$

So they ARE equivalent (perhaps not clearly so).

$\endgroup$
2
$\begingroup$

When you check your work with Wolfram Alpha, you can also use WA to find out if the form it gives is actually equivalent to yours.

Simply write your answer (without the constant), then subtract WA's answer (without the constant). If the result is a constant, the answers are equivalent.

In your example, WA says the difference is $-\frac\pi2,$ which tells you you did OK:

https://www.wolframalpha.com/input/?i=%28-arccos%28x%2F5%29%29+-+arcsin%28x%2F5%29

$\endgroup$
1
$\begingroup$

Depends on how you differentiate $\dfrac{dy}{dx}(\cos^{-1}{\dfrac{x}{5}})$

$y=\cos^{-1}{\dfrac{x}{5}}$

$x=5\cos{y}$

$\dfrac{dx}{dy}=-5\sin{y}=-(\sqrt{25-25\cos^2{y}})=-(\sqrt{25-(5\cos{y})^2})=-(\sqrt{25-x^2})$

$\dfrac{dy}{dx}=\dfrac{1}{-\sqrt{25-x^2}}$

$=\dfrac{dy}{dx}=\dfrac{1}{\sqrt{25-x^2}} \text{I took out the negative sign in the beginning because I was getting confused}$

$\endgroup$

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.