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