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.

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