Question

ayuda con esta integral ∫sen^4 x.dx

Answer 1

[sin^2(x)]^2

=>[(1 -cos(2x)/2)]^2

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

=>1/4[1 - 2cos(2x) + (1 + cos(4x)/2 ] se usa la identidad para cos^2x

=>(1/4) - (1/2)cos(2x) + (1/8) + (1/8)cos(4x) cambio de variable para el angulo

=>(3/8) - (1/2)cos(2x) + (1/8)cos(4x)

 ∫sin^4(x)dx = 3/8∫dx - 1/2∫cos(2x) dx + 1/8∫cos(4x) dx

∫sin^4(x)dx = (3/8)x - (1/2)sin(2x)*(1/2) + (1/8)sin(4x)*(1/4) + c

∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + c