Question

Derivation and application of reduction formula?

1. "Use integration by parts to derive the reduction formula `\int\cos^n(x)dx=1/n\sinx\cos^(n-1)(x)+(n-1)/n\int\cos^(n-2)(x)dx`, where `n` is a positive integer."
   I assume that I need to split this into `\int\cosx\cos^(n-1)x` or something to get a valid IBP process?

2. Use the previous reduction formula or integration by parts to evaluate: `\int\cos^3dx`
   (Where is the `x` in this problem??)

Answer 1

Write the integrand as:

`cos^nx = cos^(n-1)x cosx = cos^(n-1)x d(sinx)`

then we can integrate by parts:

`int cos^nxdx = cos^(n-1)x sinx -int sinx d( cos^(n-1)x)`

`int cos^nx dx= cos^(n-1)x sinx -(n-1) int sinx cos^(n-2)x (-sinx) dx`

`int cos^nxdx = cos^(n-1)x sinx +(n-1) int sin^2 x cos^(n-2)xdx`

Now use: `sin^2x = 1-cos^2x`:

`int cos^nx dx= cos^(n-1)x sinx +(n-1) int (1-cos^2x) cos^(n-2)xdx`

and using the linearity of the integral:

`int cos^nx dx= cos^(n-1)x sinx +(n-1) int cos^(n-2)xdx - (n-1) int cos^nxdx`

The integral:

`I_n = int cos^nx dx`

appears now on both sides of the equation and we can solve for it:

`I_n = cos^(n-1)x sinx +(n-1)I_(n-2) - (n-1)I_n`

`n I_n = cos^(n-1)x sinx +(n-1)I_(n-2)`

`I_n = (cos^(n-1)x sinx)/n +(n-1)/nI_(n-2)`

which proves the reduction formula.

For `n=3` in particular we have:

`int cos^3x dx= (cos^2xsinx)/3 + 2/3 int cosxdx`

`int cos^3x dx= (cos^2xsinx+2sinx )/3 +C`

and simplifying:

`int cos^3x dx= ((1-sin^2x)sinx+2sinx )/3 +C`

`int cos^3x dx= (3sinx -sin^3x)/3 +C`

`int cos^3x dx= sinx -sin^3x/3 +C`