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# Where did I go wrong?

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$$\int \frac{d}{dx} [{-arcsin}(x)] \hspace{0.1cm} dx = -arcsin(x)$$

$$\int d [{-arcsin(x)}]=-arcsin(x)$$

$$\int -d[arcsin(x)]=-arcsin(x)$$

$$\int -\bigg[\frac{1}{\sqrt{1-x^2}}\bigg] = -arcsin(x)$$

You could pull the negative out and this equation is clearly true... or

$$arccos(x)=-arcsin(x)$$

Why is that identity not true? Where did I go wrong?

Mar 2, 2019

#1
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$$\displaystyle \int - \left[\dfrac{1}{\sqrt{1-x^2}}\right]~dx = -\arcsin(x) \text{, not }\arccos(x)\text{ for one}$$

.
Mar 2, 2019
#2
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Wait, so what is the derivative of arccos(x)?

Guest Mar 2, 2019
#3
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Rom could you check this please.

I do not know thise off by heart, I have to work them out every time

$$If\;\;\; \theta=acos(x)\quad then\;\;\;find \; \frac{d\theta}{dx}\\~\\ x=cos\theta\\ \frac{dx}{d\theta}=-sin\theta\\$$

Now we know that  cos(theta) =x

so I draw a right angled triangle  where this is true.

$$\text{adjacent=x hypotenuse=1 opposite=} \sqrt{1-x^2}$$

$$\frac{dx}{d\theta}=-sin\theta\\ \frac{d\theta}{dx}=\frac{-1}{\sqrt{1-x^2}}\\$$

So it seems to me that

$$\displaystyle \int - \left[\dfrac{1}{\sqrt{1-x^2}}\right]~dx = \arccos(x) +c$$

Wolfram|alpha seems to agree with me.

Melody  Mar 2, 2019
#4
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Hmm!   -arcsin(x) is the same as arccos(x) to within a constant, so either is acceptable in a solution to the indefinite integral. Mar 3, 2019
#5
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Thanks Alan.

In other works Guest the identity you got is true except that you forgot to include the constant.

Yes I know you already said that Alan :)

Melody  Mar 4, 2019
edited by Melody  Mar 4, 2019
edited by Melody  Mar 4, 2019