Is Sin^-1 (x) = - Cos^-1 (x)

Question

In integration, it is given as:

 

Integration of 1/(1-x^2)^(1/2) = sin^-1 (x)

 

and

 

Integration of 1/(1-x^2)^(1/2) = - cos^-1 (x)

 

Does this imply both are equal?

Answer 1

With those two formulas you have forgotten to include a constant of integration!

 

∫dx/√[1-x²] = arcsin(x) + C₁

*or*

∫dx/√[1-x2] = -arccos(x) +C₂

 

C₁ and C₂ are not necessarily equal to each other. When two functions are possible solutions to an (indefinite) integral, the most we can say is that the functions will differ by a constant. That means there is some number C₃ that satisfies

 

arcsin(x)=-arccos(x)+C₃

 

And in fact, if you look up a list of identities for inverse trig functions, you will eventually come across

 

arcsin(x)=π/2 - arccos(x)

 

If you take the derivative of both sides of the above:

 

1/√[1-x²] = 0 - (-1/√[1-x2])

1/√[1-x2] = 1/√[1-x2] 

 

 

Answer 2

d/dx(Cos^-1x) = -d/dx(Sin^-1x), but Cos^-1x ≠ -Sin^-1x.

 

For example, Cos^-1(1) = 0, but Sin^-1(1) = π/2.

 

   So, Cos^-1x ≠ -Sin^-1x