Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 5

You need to re-write the expression t^2 - 1 , such that:
t^2 - 1 = t^2(1 - (1/t)^2)
You need to use the following substitution, such that:
1/t = sin u => -1/(t^2) dt = cos u du => (dt)/(t^2) = -cos u du
u = arcsin (1/t)
int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) = - int_(u_1)^(u_2) (cos u du)/(1/(sin u)*sqrt(1 - sin^2 u))
You need to use the basic trigonometric formula 1 - sin^2 u = cos^2 u, such that:
- int_(u_1)^(u_2) (cos u*sin u du)/(sqrt(1 - sin^2 u)) = - int_(u_1)^(u_2)
- int_(u_1)^(u_2) (cos u*sin u du)/(sqrt(cos^2 u)) = - int_(u_1)^(u_2) (cos u*sin u du)/(cos u)
Reducing like terms yields:
-int_(u_1)^(u_2) (sin udu) = -(-cos u)|_(u_1)^(u_2)
Replacing back the variable, yields:
int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) =(cos (arcsin (1/t)))_(sqrt 2)^2
int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) =cos (arcsin (1/2)) - cos (arcsin (1/sqrt 2))
int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) = cos(pi/6) - cos(pi/4)
int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) = sqrt3/2 - sqrt2/2
int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) = (sqrt3 - sqrt2)/2
Hence, evaluating the given integral yields int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) = (sqrt3 - sqrt2)/2.

Comments

Popular posts from this blog

How does Bilbo show leadership and courage in The Hobbit?

In “Goodbye to All That,” Joan Didion writes that the “lesson” of her story is that “it is distinctly possible to remain too long at the fair.” What does she mean? How does the final section of the essay portray how she came to this understanding, her feelings about it, and the consequences of it?

Why does the poet say "all the men and women merely players"?