this `int (1+x^(-2))^(1/2) dx=int sqrt(1+x^(-2)) dx` which can be written as:
`intsqrt((1+1/x^2))dx`
we can simplify the `1+1/x^2` by taking `x^2` as common denominator which becomes `(x^2+1)/x^2` which looks:
`sqrt(((x^2+1)/x^2))=(sqrt(x^2+1))/(sqrt(x^2))=(sqrt(x^2+1))/x` now our simplified equation looks:
`int(sqrt(x^2+1))/xdx`
Now because we have `sqrt(x^2+1)` in our equation, we can use trig substitution by letting
`x=tantheta`
`dx=sec^2thetad theta`
sub in the values we have: `int(sqrt(x^2+1))/xdx=int(sqrt((tantheta)^2+1))/(tan theta)sec^2thetad theta`
from trig identity we know that `1+(tantheta)^2=(sectheta)^2` so under the square root will be replaced by `(sectheta)^2` and we have:
`int(sqrt((sectheta)^2)/(tan theta)sec^2thetad theta`
The square root and root cancels out and we have:
`intsectheta/(tan theta)sec^2thetad theta` which we can write `intsectheta*1/(tan theta)sec^2thetad theta`
we know that `secx=1/cosx` so we can rewrite this as
`int1/costheta*1/(tan theta)sec^2thetad theta`
and we know that `sinx/cosx=tanx` therefore `1/tanx=cosx/sinx` so we can substitute it in our equation and we have:
`int1/costheta*costheta/sintheta sec^2thetad theta`
Now the `costheta` at the top and bottom cancel out and we have
`int1/sintheta sec^2thetad theta`
`1/sinx=cscx` and applying this in our equation we have:
`intcsctheta *sec^2thetad theta`
so far we have simplified our equation very good. We know `1+tan^2x=sec^2x` applying this to our equation we have:
`intcsctheta (1+tan^2theta)d theta`
Multiplying `csctheta` to inside we have:
`intcsctheta +cscthetatan^2thetad theta`
At this stage have + sign in our equation and we can split it to separate integrals as follow:
`intcscthetad theta +intcscthetatan^2thetad theta`
Part 1for the first part `intcscthetad theta=ln|csctheta-cottheta|+C`
Now for the `intcscthetatan^2thetad theta` we can simplify it by `(sin^2x)/(cos^2x)=tan^2x` :
`intcsctheta*(sin^2theta)/(cos^2theta)d theta`
we know `cscx=1/sinx` applying this to our equation we have:
`int1/sintheta*(sin^2theta)/(cos^2theta)d theta` and one of the top `sintheta` cancels out with the bottom one and have:
`int(sintheta)/(cos^2theta)d theta`
Letting `u=(costheta)^2`
`du=-2sinthetad theta` multiply both side by `1/-2` we get `1/-2du=sinthetad theta`. Now sub these values into our equation we have:
`int1/(-2*u^2)du=1/-2int1/u^2du`
integrate we have `ln|u|+C` we had `u=(costheta)^2` so our equation `ln|u|+C=ln|(costheta)^2|+C`
Putting them from Part 1 we had `ln|csctheta-cottheta|+C` we have:
`ln|csctheta-cottheta|+ln|(costheta)^2|+C`
Because our equation was in terms of x, so we have now in terms of
so we have to convert our equation using right angel triangle.
we initially set `x=tantheta` . We know when `thantheta=x` which is actually `thantheta=x/1` then we found two side of triangle therefor we have:
`h^2=x^2+1^2` according to Praetorian theorem the h is hypotenuses so `h=sqrt( x^2+1^2)=sqrt( x^2+1)`
All fo the identities are as follow:
`sintheta=x/sqrt( x^2+1)` so `csctheta=sqrt( x^2+1)/x`
`costheta=1/sqrt( x^2+1)` so `sectheta=sqrt( x^2+1)`
`thantheta=x` so `cottheta=1/x`
using the above values we can replace them
in `ln|csctheta-cottheta|+ln|(costheta)^2|+C` replace `costheta` we have:
in `ln|csctheta-cottheta|+ln|(1/sqrt( x^2+1))^2|+C`
Replace `csctheta=( x^2+1)/x` and `cottheta=1/x` we have:
in `ln|sqrt( x^2+1)/x-1/x|+ln|(1/sqrt( x^2+1))^2|+C`
and that is the final answer
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