Help on the easy integration?

Question

integral of (sinx)^4*(cosx)^2 dx

i have no idea where to start..

thx

Answer 1

sin^4 x * cos^2 x

= (1/8) * (2sin^2 x) * (4sin^2 x cos^2 x)

= (1/8) * (1 - cos 2x) * (sin^2 2x)

= (1/16) * (1 - cos 2x) * (2sin^2 2x)

= (1/16) * (1 - cos 2x) * (1 - cos 4x)

= (1/16) * (1 - cos 2x - cos 4x + cos 2x cos 4x)

= (1/32) * (2 - 2cos 2x - 2cos 4x + 2cos 2x cos 4x)

= (1/32) * (2 - 2cos 2x - 2cos 4x + cos 6x + cos 2x)

= (1/32) * (2 - cos 2x - 2cos 4x + cos 6x)

=> ∫sin^4 x * cos^2 x dx

= (1/32) ∫(2 - cos 2x - 2cos 4x + cos 6x) dx

= (1/32) [2x - (1/2)sin 2x - (1/2) sin 4x + (1/6) sin 6x) + c.

Answer 2

use these conversions:

cos²x = (1 + cos(2x))/2

sin²x = (1 - cos(2x))/2

thus

∫ (sin x)^4 (cos x)^2 dx

= ∫ [(1 - cos(2x))^2 / 4] [(1 + cos(2x))/2] dx

= 1/8 ∫ [{cos(2x)}^3 - {cos(2x)}^2 - cos(2x) + 1] dx

i will add the "+C" in the end .. . .

first integral:

∫ {cos(2x)}^3 dx = ∫ (1 - sin²(2x)) (cos 2x) dx

now, let u = sin(2x) , du = 2cos(2x) dx

integral = 1/2 ∫ (1 - u^2) du = 1/2 (u - u^3/3)

= 1/2 (sin(2x) - {sin(2x)}^3/3)

second integral:

∫ -{cos(2x)}^2 dx = ∫ -[1 + cos(4x)]/2 dx

= -1/2 x - 1/8 sin(4x)

last portion of integrals:

∫ [-cos(2x) + 1] dx = -1/2 sin(2x) + x

thus .. . .. . ∫ (sin x)^4 (cos x)^2 dx

= 1/2 (sin(2x) - {sin(2x)}^3/3) - 1/2 x - 1/8 sin(4x) -1/2 sin(2x) + x + C

= -{sin(2x)}^3/3 - 1/8 sin(4x) + 1/2 x + C