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

Question

Isaiah Allen · Accepted Answer

#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#

Aurora Alvarado · Answer

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

Have a look:

Luke Alvarez · Answer

#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#

Elijah Abram · Answer

undefined To integrate $ \int \cos^3(x) \, dx $, you can use the trigonometric identity $ \cos^2(x) = 1 - \sin^2(x) $. Then, perform a substitution by letting $ u = \sin(x) $. This will change $ du = \cos(x) \, dx $. With these substitutions, the integral becomes $ \int (1 - u^2) \, du $. Integrating term by term, you get $ u - \frac{u^3}{3} + C $, where $ C $ is the constant of integration. Finally, revert back to the variable $ x $ using $ u = \sin(x) $. Thus, the solution is $ \sin(x) - \frac{\sin^3(x)}{3} + C $.