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??)

1 Answer
Apr 9, 2018

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#