Reduction formulas for integral of sin and cos

[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 I am not seeing it for some reason.

 

[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)​