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

int1/(t^2sqrt(t^2-16))dt

Let's evaluate the integral by applying integral substitution,

Let t=4sec(u)

=>dt=4tan(u)sec(u)du

Plug the above in the integral,

int1/(t^2sqrt(t^2-16))dt=int(4tan(u)sec(u))/((4^2sec^2(u)sqrt(4^2sec^2(u)-16)))du

=int(4tan(u)sec(u))/(16sec^2(u)*4sqrt(sec^2(u)-1))du

=inttan(u)/(16sec(u)sqrt(sec^2(u)-1))du

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

=1/16inttan(u)/(sec(u)sqrt(1+tan^2(u)-1))du

=1/16inttan(u)/(sec(u)tan(u))du

=1/16int1/sec(u)du

=1/16intcos(u)du

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

=1/16sin(u)+C

We have used t=4sec(u)

=>u=arcsec(t/4)

Substitute back u in the solution,

=1/16sin(arcsec(t/4))+C

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,

4^2+O^2=t^2

O^2=t^2-16

O=sqrt(t^2-16)

sin(theta)=sqrt(t^2-16)/t

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

Approved by eNotes Editorial Team

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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