Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 17

Indefinite integral are written in the form of int f(x) dx = F(x) +C
where: f(x) as the integrand
F(x) as the anti-derivative function
C as the arbitrary constant known as constant of integration
For the given problem int xarcsec(x^2+1) dx, it has a integrand in a form of inverse secant function. The integral resembles one of the formulas from the integration as : int arcsec (u/a)du = u*arcsin(u/a) +-aln(u+sqrt(u^2-a^2))+C .
where we use: (+) if 0ltarcsec (u/a)ltpi/2
(-) if pi/2ltarcsec(u/a)ltpi
Selecting the sign between (+) and (-) will be crucial when solving for definite integral with given boundary values [a,b] .
For easier comparison, we may apply u-substitution by letting:
u =x^2+1 then du = 2x dx or (du)/2
Plug-in the values int xarcsec(x^2+1) dx , we get:
int xarcsec(x^2+1) dx=int arcsec(x^2+1) * xdx
= int arcsec(u) * (du)/2
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int arcsec(u) * (du)/2= 1/2int arcsec(u) du
or 1/2 int arcsec(u/1) du
Applying the aforementioned formula from the integration table, we get:
1/2 int arcsec(u/1) du=1/2 *[u*arcsin(u/1) +-1ln(u+sqrt(u^2-1^2))]+C
=1/2 *[u*arcsin(u) +-ln(u+sqrt(u^2-1))]+C
=(u*arcsin(u))/2 +-(ln(u+sqrt(u^2-1)))/2+C
Plug-in u =x^2+1 on (u*arcsin(u))/2 +-(ln(u+sqrt(u^2-1)))/2+C , we get the indefinite integral as:
int xarcsec(x^2+1) dx=((x^2+1)*arcsin(x^2+1))/2 +-(ln(x^2+1+sqrt((x^2+1)^2-1)))/2+C
=(x^2arcsin(x^2+1))/2+arcsin(x^2+1)/2 +-ln(x^2+1+sqrt(x^4+2x^2))/2+C
=(x^2arcsin(x^2+1))/2+arcsin(x^2+1)/2 +-ln(x^2+1+sqrt(x^2(x^2+2)))/2+C
=(x^2arcsin(x^2+1))/2+arcsin(x^2+1)/2 +-ln(x^2+1+|x|sqrt(x^2+2))/2+C

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