Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 8

You need to re-write the expression t^2 - 16 , such that:
t^2 - 16 = t^2(1 - (4/t)^2)
You need to use the following substitution, such that:
4/t = sin u => -4/(t^2) dt = cos u du => (dt)/(t^2) = -(1/4)*cos u du
u = arcsin (4/t)
int (dt)/(t^2*sqrt(t^2-16)) = (-1/4) int (cos u du)/(4/(sin u)*sqrt(1 - sin^2 u))
You need to use the basic trigonometric formula 1 - sin^2 u = cos^2 u , such that:
(-1/16) int (cos u*sin u du)/(sqrt(1 - sin^2 u)) = (-1/16) int (cos u*sin u du)/(sqrt(cos^2 u))
(-1/16) int (cos u*sin udu)/(sqrt(cos^2 u)) = (-1/16) int (cos u*sin udu)/(cos u)
Reducing like terms yields:
(-1/16) int (sin udu) = (-1/16)(-cos u) + c
Replacing back the variable, yields:
int (dt)/(t^2*sqrt(t^2-16)) = (1/16)(cos (arcsin (4/t))) + c
Hence, evaluating the given integral yields int (dt)/(t^2*sqrt(t^2-16)) = (1/16)(cos (arcsin (4/t))) + c.

Comments

Popular posts from this blog

In “Fahrenheit 451,” what does Faber mean by “Those who don’t build must burn. It’s as old as history and juvenile delinquents”?

Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 34

What was the effect of World War II on African Americans?