3 Questions on Inverse, Reciprocal, and Trigonometric Integration

[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

 

[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 ;)

 

[HallsofIvy]

Hi again everyone…

I have 3 questions:

1) What is the difference between INVERSE and RECIPROCAL?

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.

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

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

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

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).