`int (dt)/(t^2 sqrt(t^2 - 16))` Evaluate the integral

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Let's evaluate the integral by applying integral substitution,

Let `t=4sec(u)`


Plug the above in the integral,




Now use the identity:`sec^2(x)=1+tan^2(x)`





Now use the standard integral:`intcos(x)dx=sin(x)+C`


We have used `t=4sec(u)`


Substitute back u in the solution,


Simplify the above by assuming the right triangle with angle `theta=arcsec(t/4)`

Hypotenuse is t and adjacent side is 4, Opposite side(O) can be found by using pythagorean identity,





Hence the solution is `1/16(sqrt(t^2-16)/t)+C`


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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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