Prove the following reduction formula: integrate of (tan^(n)x) dx= (tan^(n-1)x)/(n-1) - integrate of (tan^(n-2))dx
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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`
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`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`