- #1
livvy07
- 6
- 0
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.
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.