How do you integrate #int cos^3x dx#?

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Answer 1 · 2016-08-04T13:08:07

I found: #sin(x)-(sin^3(x))/3+c#

Have a look:
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Answer 2 · 2016-08-04T15:06:15

#sin(x)cos(x)^2+2/3sin(x)^3+C#

#d/(dx)(sin(x)cos(x)^2)=cos(x)^3-2sin(x)^2cos(x)#

then

#int cos(x)^3dx = sin(x)cos(x)^2+2intsin(x)^2cos(x)dx#

but

#intsin(x)^2cos(x)dx = int(1/3 d/(dx)sin(x)^3)dx#

finally

#int cos(x)^3dx = sin(x)cos(x)^2+2/3sin(x)^3+C#

Answer 3 · 2016-08-04T22:02:44

#int cos^3 x d x=sin x-1/3sin^3 x+C#

#"a different way..."#

#"use reduction formula"#

#int cos^n x d x=(n-1)/n int cos^(n-2) x d x+(cos^(n-1)x*sin x)/n#

#"use n=3"#

#int cos^3 x d x=(3-1)/3 int cos^(3-2)x +(cos^(3-1)x*sin x)/(3)#

#int cos^3 x d x=2/3int cos x d x+(cos^2 x*sin x)/3#

#cos^2x=1-sin^2x#

#int cos^3 x d x=2/3 sin x+((1-sin^2 x)*sin x)/3#

#int cos^3 x d x=2/3 sin x+(sin x-sin^3 x)/3#

#int cos^3 x d x=(2 sin x+sin x-sin^3 x)/3#

#int cos^3 x d x=(3sin x-sin^3 x)/3#

#int cos^3 x d x=sin x-1/3sin^3 x+C#