integral of (sinx)^4*(cotx)^3
help please.
cotx = cosx/sinx
y = integral of (sinx)^4*(cotx)^3
= integral of (sinx)^4*(cosx)^3/(sinx)^3
= integral of sinx*(cosx)^3
let cosx = u
sinxdx = du
so, y = integral of u^3 du
= (u^4)/4
= (1/4)(cosx)^4 + constant
(sinx)^4*(cotx)^3
= sin^4(x) * [cos^3(x) / sin^3(x)]
= cos^3(x) sin(x)
Let u = cos^3x
du = -sinx dx
-1∫ cos^3(x) (-sin(x)) dx
= -1/4 cos^4x + c
Copyright © 2024 Q2A.ES - All rights reserved.
Answers & Comments
Verified answer
cotx = cosx/sinx
y = integral of (sinx)^4*(cotx)^3
= integral of (sinx)^4*(cosx)^3/(sinx)^3
= integral of sinx*(cosx)^3
let cosx = u
sinxdx = du
so, y = integral of u^3 du
= (u^4)/4
= (1/4)(cosx)^4 + constant
(sinx)^4*(cotx)^3
= sin^4(x) * [cos^3(x) / sin^3(x)]
= cos^3(x) sin(x)
Let u = cos^3x
du = -sinx dx
-1∫ cos^3(x) (-sin(x)) dx
= -1/4 cos^4x + c