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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
avatar+5225 
+1

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

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

Guest Mar 2, 2019
 #3
avatar+102447 
+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
 #4
avatar+28064 
+4

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
avatar+102447 
0

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

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