Question

Evaluate the following: `int_(pi/6)^(pi/2)(cscxcotx)dx` I know that we are suppose to find the anti-derivative of cscxcotx, but i dont know how to?

Answer 1

Recall that `d/dxcscx=-cscxcotx`. Reversing this shows that `int(-cscxcotx)dx=cscx+C`.

So, we have

`int_(pi//6)^(pi//2)(cscxcotx)dx=-int_(pi//6)^(pi//2)(-cscxcotx)dx=[-cscx]_(pi//6)^(pi//2)`

So, the anti-derivative you were looking for was `-cscx`. The actual evaluation of the integral gives:

`-csc(pi/2)-(-csc(pi/6))=-1+2=1`

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Another way to find the integral is to use `cscx=1/sinx` and `cotx=cosx/sinx`:

`int(cscxcotx)dx=int1/sinxcosx/sinxdx=intcosx/sin^2xdx`

This can be found with the substitution `u=sinx`, which implies that `du=cosxdx`.

`=int1/u^2du=intu^-2du`

Using `intu^ndu=u^(n+1)/(n+1)`:

`=u^-1/(-1)=-1/u=-1/sinx=-cscx+C`

As we determined above!