what is the integration of cos^2m (x).?

Answer 1

[Snoopy: {e^jx + e^(-jx)}^2m <> {e^(j2mx) + e(-j2mx)} - Binomial Theorem says it is e^(j.2mx) + (2m)e^(jx(2m-2)) + ... + C(2m, k) e^(jx(2m-2k)) + ... + e^(-j2mx).]


Using cos 2x = cos^2 x - 1 we rewrite cos^2m x in terms of cos 2x:
cos^(2m) x
= (cos^2 x)^m
= ((cos 2x + 1)/2)^m
= 2^(-m) (cos^m 2x + m cos^(m-1) 2x + ... + 1)
So we have a polynomial in cos 2x. Odd-powered terms we deal with by separating out one power and substituting u = sin 2x, du = 2 cos 2x:
∫cos^(2k+1) 2x
= ∫(cos 2x) (1 - sin^2 2x)^k dx
= (1/2) ∫(1-u^2)^k du
For even powered terms we repeat the above process, halving the degree of the polynomial at each step and expressing it in terms of cos 4x, cos 8x, etc.

Example 1: ∫cos^6 x dx
= ∫(cos^2 x)^3 dx
= (1/8) ∫(cos 2x + 1)^3 dx
= (1/8) ∫(cos^3 2x + 3 cos^2 2x + 3 cos 2x + 1) dx
= 1/8 [∫(1 - sin^2 2x) cos 2x dx + 3∫(cos 4x + 1)/2 dx + 3/2 sin 2x + x]
= 1/8 [∫(1 - u^2) / 2 du + 3/2 (sin 4x / 4 + x) + 3/2 sin 2x + x]
= 1/8 [(u/2 - u^3/6) + 5/2 x + 3/8 sin 4x + 3/2 sin 2x] + c
= 1/8 [2 sin 2x - 1/6 sin^3 2x + 3/8 sin 4x + 5/2 x] + c
= 1/4 sin 2x - 1/48 sin^3 2x + 3/64 sin 4x + 5/16 x + c.

Example 2: ∫cos^8 x dx
= 1/16 ∫(cos 2x + 1)^4 dx
= 1/16 ∫(cos^4 2x + 4 cos^3 2x + 6 cos^2 2x + 4 cos 2x + 1) dx
= 1/64 ∫(cos 4x + 1)^2 dx + 1/16 ∫(4 cos 2x) (cos^2 2x + 1) dx + 3/16 ∫(cos 4x + 1) dx + 1/16 x
= 1/64 ∫(cos^2 4x + 2 cos 4x + 1) dx + 1/16 ∫(4 cos 2x) (2 - sin^2 2x) dx + 3/16 ∫(cos 4x + 1) dx + 1/16 x
= 1/128 ∫(cos 8x+ 1) dx + 7/32 ∫cos 4x dx + 17/64 x + 1/16 ∫(2 du) (2 - u^2)
= 1/128 (sin 8x / 8) + 1/128 x + 7/32 (sin 4x / 4) + 17/64 x + 1/16 (4u - 2/3 u^3) + c
= 1/1024 sin 8x + 7/128 sin 4x + 35/128 x + 1/4 sin 2x - 1/24 sin^3 2x + c.

Answer 2

Yes: Scarlet Manuka is correct. Thanks ..
Integral Cos ^2m(x)=Integral 0.5*{e^jx + e^(-jx)}^2m
=Integral 0.5*{e^(j2mx) + e(-j2mx)}
=0.5*{e^(j2mx)}/2m - {e^(-j2mx)}/2m + C