int cos^n x dx=(1/n)cos^(n-1)x sin x + (n-1)/n int cos^(n-2)x dx∫cosnxdx=(1n)cosn−1xsinx+n−1n∫cosn−2xdx.
To derive the reduction formula, rewrite cos^n xcosnx as cos x cos^(n-1)xcosxcosn−1x and then integrate by parts.
Let I_nIn denote int cos^n x dx∫cosnxdx
I_n=sin x cos^(n-1)x - int (sin x )(n-1)cos^(n-1)x(-sin x)dxIn=sinxcosn−1x−∫(sinx)(n−1)cosn−1x(−sinx)dx
followed by using sin^2x=1-cos^2xsin2x=1−cos2x to get the sin^2sin2 back to a cosine
But this gives you (n-1)int cos^nx dx(n−1)∫cosnxdx somewhere on the right:
I_n=sin x cos^(n-1)x + (n-1)I_(n-2)-(n-1)I_nIn=sinxcosn−1x+(n−1)In−2−(n−1)In.
Do not panic that you appear to have got back to I_nIn! All you need to do is cancel the I_nIns and move the -nI_n−nIn to the left hand side:
n int cos^n x dx=sin x cos^(n-1)x + (n-1) int cos^(n-2)x dxn∫cosnxdx=sinxcosn−1x+(n−1)∫cosn−2xdx .
Dividing through by nn gives the reduction formula.
For any particular small positive value of nn you can apply this repeatedly to get down to the integral either of 11 or of cos xcosx.
