Integrating cos^-2(x) after inverse substitution

[mathtrickedme]

Homework Statement

Use the substitution x=4sin(t) to evaluate the integral: S 1/[(16-x^2)^(3/2)] dx

Homework Equations

x = 4sin(t)

The Attempt at a Solution

x = 4sin(t)
dx = 4cos(t) dt

4cos(t) = (16-x^2)^(1/2), i cube both sides to get

(4cos(t))^3 = (16-x^2)^(3/2), then plug in dx and denominator into the equation to get

S 4cos(t)/[(4cos(t))^3] dt simplified and constant taken out i now get
(1/16) S [cos^-2(t)] dt

how do i integrate cos^-2(t) to get a simple answer. I don't think i can use the S cos^n(t) formula because then i have to integrate sin^-4(t) afterwards.

any help would be awesome, thanks

 

[Vid]

cos^-2(x) = sec^2(x)

Integral of sec^2(x) = tan(x)