Derivative and integral (confusing part)

[daivinhtran]

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)

 

[daivinhtran]

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

 

[haruspex]

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

 

[daivinhtran]

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

Do you mean that both my and my teacher answers are right?

 

[haruspex]

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

 

[HallsofIvy]

Do you mean that both my and my teacher answers are right?

Wouldn't that be terrible!

 

[sahil_time]

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 :)