we need to show that `inttan^n(x)dx=(tan^(n-1)(x))/(n-1)-inttan^(n-2)(x)dx`

We know that `tan^2(x)=sec^2(x)-1`

hence we can rewrite our integral in the following manner:

`inttan^n(x)dx=inttan^(n-2)(x)*tan^2(x)dx=`

`inttan^(n-2)(x)(sec^2(x)-1)dx=`

`int(tan^(n-2)(x)sec^2(x)-tan^(n-2)(x))dx=`

`inttan^(n-2)(x)sec^2(x)dx-inttan^(n-2)dx=`

In order to integrate the first part let u=tan(x) then du=sec^2(x)dx

Thus the above problem becomes

`intu^(n-2)du-inttan^(n-2)(x)dx=`

`[u^(n-1)]/(n-1)-inttan^(n-2)(x)dx=`

`[tan^(n-1)(x)]/(n-1)-inttan^(n-2)(x)dx`

### Videos

`int "tan"^n x dx =`

`int ("tan" ^(n-2) x) ("tan" ^2 x) dx =`

`int ("tan" ^(n-2) x) ("sec"^2 x - 1)dx = `

`int ("tan" ^(n-2) x ) "sec"^2 x dx - int "tan" ^(n-2) x dx=`

To do the first integral, use a u substitution:

`u="tan" x` `du = "sec"^2 x dx`

`int ("tan" ^(n-2) x )"sec"^2 x dx = `

`int u^(n-2) du =`

`1/(n-1) u^(n-1) =`

`1/(n-1) "tan" ^(n-1) x`

Putting the pieces together, we get the reduction formula:

`int "tan"^n x dx = 1/(n-1) "tan" ^(n-1) x - int "tan" ^(n-2) x dx`