# Reduction formulas for integral of sin and cos

1. Feb 12, 2012

### livvy07

Integrate sin2(x)cos4(x)dx using reduction formulas?
My book says integral sin2(x)cos4(x)dx= integral cos4(x)-integral cos6x dx
Now the reduction formula for n=6
for integral cos6(x)dx= (1/6)cos5(x)sinx+(5/6) integral cos4(x)dx

Here is the part I don't get: It then says :
sin2(x)cos4(x)dx
=(-1/6)cos5(x)sinx+(1/6) integral cos^4(x)dx

I don't get how the 5/6 becomes 1/6 or why 1/6 becomes -1/6 if that makes any sense? Any help would be great! I've been looking at it for awhile, but Im not seeing it for some reason.

2. Feb 12, 2012

### tiny-tim

hi livvy07!

(have an integral: ∫ )

i don't understand that

∫ sin2(x)cos4(x) dx

= ∫ [sin(x)][sin(x)cos4(x)] dx

then integrate by parts (integrating the second bracket, and differentiating the first bracket)​