`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...
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`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`
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.`