# Prove the following reduction formula: integrate of ((cosx)^n) dx =1/n(cos^(n-1)x)(sin(x)) + ((n-1)/n) integrate of cos^(n-2)dx

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### 2 Answers

Use integration by parts:

`u="cos" ^(n-1) x` `dv = "cos" x`

`du = (n-1)( "cos"^(n-2) x)( -"sin" x) dx` `v="sin" x`

Then

`int "cos" ^n x dx = "cos"^(n-1) x "sin" x + (n-1) int "cos"^(n-2) x "sin" ^2 x dx`

`="cos"^(n-1) x "sin" x + (n-1) int "cos"^(n-2) x (1-"cos"^2 x ) dx`

`="cos"^(n-1) x "sin" x + (n-1) int "cos"^(n-2) x dx - (n-1) int "cos"^n x dx`

Notice that we have a `int "cos"^n x dx` term on both sides of the equation:

`int "cos" ^n x dx =[ ...] + [...] - (n-1) int "cos"^n x dx`

Adding `(n-1) int "cos" ^n x dx` to both sides gives:

`n int "cos" ^n x dx = "cos" ^(n-1) x "sin" x +(n-1) int "cos" ^(n-2) x dx`

Divide by n to get your reduction formula:

`int "cos" ^n x dx = (1/n) "cos" ^(n-1) x "sin" x + ((n-1)/n) \int "cos"^(n-2) x dx`

http://www.enotes.com/math/q-and-a/prove-following-reduction-formula-integrate-cosx-n-364527