integrate sin^7xdx

Answer 1

if
sin^6(x)sin(x)dx...(1)
then
let m=cos(x)
dm/dx=-sin(x)
dm=-sin(x)dx
input 1 into 2
(1-cos^2(x)^3)sin(x)dx
(1-u^2)^3-du
-(1-u^2)^3du
-[1-3u^2+3u^4-u^6)
[-1+3u^2-3u^4+u^7)
now just integrate and input your 'm'

Answer 2

thanks you

Answer 3

let sin^7x =(sin^2(x))^3sin(x) dx.
sin^6x= (sin^2(x))^3
integral of(sin^2(x))^3sinx
using identity sin^2x= 1-cos^2x
integral of (1-cos^2x)^3sinx dx
what if u=cosx
du=-sinx.
-integral(-u^6+3u^4-3u^2+1)du.
=-(-u^7/7+3u^5/5-u^3+u)+k
substituting for u give us the following
-(-cosx^7x/7+3cox^5/5-cos^3x+cosx)+k

Answer 4

sorry i made a typo i mean u not m