Evaluate ∫sin^5 xdx?

Answer 1

as
sin^5 x =sin x * sin^4 x = sinx * (1-con^2 x)^2
=sinx * (1-2 con^2 x +con^4 x)


~~ since dcosx = (-1) sinx dx
we have
integ sin ^5 xdx =integ (sinx * (1-2*con^2 x +con^4 x) dx
=integ (1-2*con^2 x + con ^4 x)(-1) dcos x
set cos x = t
original equation becomes
integ sin ^5 x dx = integ (-1)*(1-2t^2+t^4 )dt
=(-1) [t -2*(1/3) t^3 + (1/5)* t^5]
=-t +(2/3)*t^3 -1/5*t^5 where t=cosx

Answer 2

Evaluate ∫sin^5 x dx.

∫(sin^5 x)dx = ∫(sin x)(sin^4 x)dx = ∫(sin x)(sin² x)²dx
= ∫(sin x)(1 - cos² x)²dx

Let
u = cos x
du = -sin x dx

∫(sin x)(1 - cos² x)²dx = ∫(1 - u²)²(-du) = -∫(1 - 2u² + u^4)du
= -u + (2/3)u³ - u^5/5 + C
= -cos x + (2/3)cos³ x - (cos^5 x)/5 + C

Answer 3

5sin^4(x)cos(x)