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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?

Guest Mar 2, 2019

#1**+1 **

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

.Rom Mar 2, 2019

#3**+1 **

**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