does anybody help to find integral of Cos^3?

Answer 1

∫cos³θdθ
= ∫cos²θcosθdθ
= ∫(1 - sin²θ)cosθdθ
= ∫cosθ - sin²θcosθdθ
= ∫cosθdθ - ∫sin²θcosθdθ

∫cosθdθ = sinθ + c

∫sin²θcosθdθ
Let u = sinθ
du = cosθdθ
∫sin²θcosθdθ
= ∫u²du
= u³/3
= (1/3)sin³θ

∫cos³θdθ
= sinθ - (1/3)sin³θ + c

Answer 2

I'm assuming you mean (Cos x)^3

(1/3)(Sin x)[(Cos x)^2 + 2]

This is taken from a book

Answer 3

cos^2 x sin x/3 + sin x

Answer 4

cosx^3dx=(1-sinx^2)cosxdx
u=sinx du=cosxdx
so,
integral of (1-u^2)du= u-(u^3)/3
so integral of cosx^3=sinx-(sinx^3)/3

Answer 5

[(intergral cos^3 x)]
= [intergral (cos^2 x)(cos x)]
= (cos^2 x)(sin x) - [intergral (sin x)(2)(cos x)(-sin x)]
= (cos^2 x)(sin x) + (2) [intergral (sin^2 x)(cos x)]
= (cos^2 x)(sin x) + (2) [intergral (1- cos^2 x)(cos1)
= (cos^2 x)(sin x) + (2) [intergral (cos x)] - (2) [intergral (cos^3 x)]

[intergral (cos^3 x)] + (2) [intergral (cos^3 x)] = (cos^2 x)(sin x) + (2)(sin x)
(3) [intergral (cos^3 x)] = (cos^2 x)(sin x) + (2)(sin x)
[intergral (cos^3 x)] = (1/3)((cos^2 x)(sin x) + (2)(sin x)) + c

hope you understand...