
#1
Dec912, 11:12 PM

P: 70

We know that d(cos^1 (x/a))/dx = 1/sqrt(a^2  x^2) (assuming a and x are positive)
So...Why the integral of 1/sqrt(a^2  x^2) is not equal to (cos^1 (x/a)) + C?????????? Instead, my teacher says it has to be (sin^1 (x/a)) + C because integral of 1/sqrt(a^2  x^2) is sin^1 (x/a) + C.....(only put the negative sign in) She doesn't really explain though...(when I ask) 



#2
Dec912, 11:24 PM

P: 70

However, when I use wolmframalpha, it gives a different solution rather than the sin^1
http://www.wolframalpha.com/input/?i=+integrate+of+1%2Fsqrt%28a^2++x^2%29+ Here the link 



#3
Dec912, 11:49 PM

Homework
Sci Advisor
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Thanks ∞
P: 9,154

Consider the possibility that the two solutions are really the same, only differing by a constant.




#4
Dec1012, 07:31 PM

P: 70

Derivative and integral (confusing part)Do you mean that both my and my teacher answers are right??? 



#5
Dec1012, 09:23 PM

Homework
Sci Advisor
HW Helper
Thanks ∞
P: 9,154

Yes. Can you see how? Hint: sin θ = cos(π/2θ)




#6
Dec1112, 08:08 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879





#7
Dec1712, 01:16 PM

P: 108

Its because cos^1(x) and sin^1(x) are related by a constant. cos^1(x) + sin^1(x) = ∏/2 So if you write ∫1/√(a^2  x^2) = cos^1 (x/a) + C I can use the relation and write ∫1/√(a^2  x^2) = ∏/2  sin^1(x/a) + C Which is equal to ∫1/√(a^2  x^2) =  sin^1(x/a) + C1 Where C1 is some arbitrary constant = C + ∏/2. So its like haruspex said, solutions differ only by a constant. Hope that Helps :) 


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