Derivation and application of reduction formula?

  1. "Use integration by parts to derive the reduction formula ∫cosn(x)dx=1nsinxcosn−1(x)+n−1n∫cosn−2(x)dx, where n is a positive integer."
    I assume that I need to split this into ∫cosxcosn−1x or something to get a valid IBP process?

  2. Use the previous reduction formula or integration by parts to evaluate: ∫cos3dx
    (Where is the x in this problem??)

1 Answer
Andrea S.
Apr 9, 2018

Write the integrand as:

cosnx=cosn−1xcosx=cosn−1xd(sinx)

then we can integrate by parts:

∫cosnxdx=cosn−1xsinx−∫sinxd(cosn−1x)

∫cosnxdx=cosn−1xsinx−(n−1)∫sinxcosn−2x(−sinx)dx

∫cosnxdx=cosn−1xsinx+(n−1)∫sin2xcosn−2xdx

Now use: sin2x=1−cos2x:

∫cosnxdx=cosn−1xsinx+(n−1)∫(1−cos2x)cosn−2xdx

and using the linearity of the integral:

∫cosnxdx=cosn−1xsinx+(n−1)∫cosn−2xdx−(n−1)∫cosnxdx

The integral:

In=∫cosnxdx

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

In=cosn−1xsinx+(n−1)In−2−(n−1)In

nIn=cosn−1xsinx+(n−1)In−2

In=cosn−1xsinxn+n−1nIn−2

which proves the reduction formula.

For n=3 in particular we have:

∫cos3xdx=cos2xsinx3+23∫cosxdx

∫cos3xdx=cos2xsinx+2sinx3+C

and simplifying:

∫cos3xdx=(1−sin2x)sinx+2sinx3+C

∫cos3xdx=3sinx−sin3x3+C

∫cos3xdx=sinx−sin3x3+C

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