The integral of cosx (sinx)^6 dx. Please show your work. Thank you :)
2006-12-02 03:42:01 · 5 answers · asked by ANON 1 in Science & Mathematics ➔ Mathematics
∫cosx (sinx)^6 dx =
u = sin x
du = cos x dx
=>
∫u^6 du = (u^7)/7 +c
=>
∫cosx (sin x)^6 dx = [(sin x)^7]/ 7 +c
₢
2006-12-02 03:57:28 · answer #1 · answered by Luiz S 7 · 0⤊ 0⤋
Substitute u=sinx. You have du=cosx*dx so
cosx (sinx)^6 dx = u^6 du. This you can integrate with you eyes closed and your hands behind your back.
2006-12-02 11:49:47 · answer #2 · answered by Anonymous · 1⤊ 0⤋
put sinx =u du =cosx dx
replace by these values you itnegrate
u^6 du the integral of u^n is u^(n+1) /n+1
here u^7/7
final ((sinx)^7)/7
2006-12-02 11:47:11 · answer #3 · answered by maussy 7 · 1⤊ 0⤋
let sin x = y
cosx dx = dy
so we have to integrate
y^6 dy
it is y^7/7 = (sin x)^7/7+C
2006-12-02 11:46:12 · answer #4 · answered by Mein Hoon Na 7 · 2⤊ 0⤋
let sinx=t
cosxdx=dt
(sin)^6xcosxdx=t^6dt
=t^7+C
=(sin)^7x+C
2006-12-02 11:50:19 · answer #5 · answered by raj 7 · 0⤊ 0⤋