Derivative and integral (confusing part)

  • Context: Undergrad 
  • Thread starter Thread starter daivinhtran
  • Start date Start date
  • Tags Tags
    Derivative Integral
Click For Summary

Discussion Overview

The discussion revolves around the relationship between derivatives and integrals of inverse trigonometric functions, specifically focusing on the integral of -1/sqrt(a^2 - x^2) and its connection to cos^-1(x/a) and sin^-1(x/a). Participants explore the reasoning behind differing solutions and the implications of constants in integration.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that the derivative of cos^-1(x/a) is -1/sqrt(a^2 - x^2) and questions why the integral of -1/sqrt(a^2 - x^2) is not equal to cos^-1(x/a) + C.
  • Another participant mentions that their teacher claims the integral should be -sin^-1(x/a) + C, citing that the integral of 1/sqrt(a^2 - x^2) is sin^-1(x/a) + C.
  • A participant suggests that both solutions might be equivalent, differing only by a constant.
  • Further discussion includes the relationship between cos^-1(x) and sin^-1(x), where it is noted that cos^-1(x) + sin^-1(x) = π/2.
  • One participant elaborates that the integral can be expressed in terms of sin^-1(x/a) by incorporating the constant relationship, leading to a conclusion that the solutions differ only by a constant.

Areas of Agreement / Disagreement

Participants express differing views on the correct form of the integral, with some suggesting that both the teacher's and the participant's answers could be valid due to the constant difference. The discussion remains unresolved regarding which interpretation is definitively correct.

Contextual Notes

There is an underlying assumption that a and x are positive, and the discussion does not resolve the mathematical steps or the implications of the constants involved in the integration process.

daivinhtran
Messages
68
Reaction score
0
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)
 
Last edited:
Physics news on Phys.org
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
 
Consider the possibility that the two solutions are really the same, only differing by a constant:wink:.
 
haruspex said:
Consider the possibility that the two solutions are really the same, only differing by a constant:wink:.



Do you mean that both my and my teacher answers are right?
 
Yes. Can you see how? Hint: sin θ = cos(π/2-θ)
 
daivinhtran said:
Do you mean that both my and my teacher answers are right?

Wouldn't that be terrible!:devil:
 
daivinhtran said:
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)
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 :)
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 29 ·
Replies
29
Views
5K