# 3 questions, help me

1. Sep 5, 2007

### tomcenjerrym

Hi again everyone…

I have 3 questions:

1) What is the difference between INVERSE and RECIPROCAL?

2) Does sin^−1 (u/a) equal to arc sin (u/a)?

3) What is the EASIEST way to remember that integral du / (√(a^2 − u^2)) = sin^−1 (u/a) + C or any transformation of trigonometric integration?

Thanks

2. Sep 5, 2007

### genneth

1) RECIPROCAL is the INVERSE to multiplication.
2) Yes
3) Just remember it... there are far more terrifying identities -- better get used to it ;)

3. Sep 5, 2007

### HallsofIvy

Staff Emeritus
It has already been pointed out that the reciprocal gives the multiplicative inverse. More generally, if f(x) is a function, its reciprocal is 1/f(x). Its inverse (if it has one) is the function f-1(x) such that f-1(f(x))= f(f-1(x))= x. The reciprocal is the inverse under the operation of multiplication, the "inverse function" is the inverse under the operation of composition of functions.

Yes, they are different notations for the same thing (arcsin(x) is just a bit old fashioned).

Remember that sin2(x)+ cos2(x)= 1 so that cos2(x)= 1- sin2(x). $\sqrt{a^2- u^2}= a \sqrt{1- (u/a)^2}$ so the substitution $sin(\theta)= u/a$ gives $cos(\theta)d\theta= (1/a)du$ and the integral becomes
$$\int \frac{du}{sqrt{a^2- u^2}}= \frac{1}{a}\int\frac{du}{\sqrt{1-(u/a)^2}}$$
$$= \int \frac{cos(\theta)d\theta}{\sqrt{1- sin^2(\theta)}}= \int d\theta= \theta+ C$$
Of course, since $sin(\theta)= u/a$, $\theta= sin^{-1}(u/a)$.