153k views
8 votes
What is Cos^-1( -sq2/2)? Pleaseee help ASAPOL

1 Answer

6 votes

Answer:

s cos^(-1)(-(q^2 s)/2) + sqrt(4 - q^4 s^2)/q^2 + constant

Explanation:

Take the integral:

integral cos^(-1)(-(q^2 s)/2) ds

For the integrand cos^(-1)(-(q^2 s)/2), substitute u = -(q^2 s)/2 and du = -q^2/2 ds:

= -2/q^2 integral cos^(-1)(u) du

For the integrand cos^(-1)(u), integrate by parts, integral f dg = f g - integral g df, where

f = cos^(-1)(u), dg = du, df = -1/sqrt(1 - u^2) du, g = u:

= -(2 u cos^(-1)(u))/q^2 + 2/q^2 integral-u/sqrt(1 - u^2) du

Factor out constants:

= -(2 u cos^(-1)(u))/q^2 - 2/q^2 integral u/sqrt(1 - u^2) du

For the integrand u/sqrt(1 - u^2), substitute p = 1 - u^2 and dp = -2 u du:

= -(2 u cos^(-1)(u))/q^2 + 1/q^2 integral1/sqrt(p) dp

The integral of 1/sqrt(p) is 2 sqrt(p):

= (2 sqrt(p))/q^2 - (2 u cos^(-1)(u))/q^2 + constant

Substitute back for p = 1 - u^2:

= (2 sqrt(1 - u^2))/q^2 - (2 u cos^(-1)(u))/q^2 + constant

Substitute back for u = -(q^2 s)/2:

Answer: = s cos^(-1)(-(q^2 s)/2) + sqrt(4 - q^4 s^2)/q^2 + constant

User Galileo
by
4.4k points