How do you find the antiderivative of #cos^3 (x)dx#?

Answer 1

Charles Anctil

# sin x- 1/3 sin^3x + C#

#\int cos^3 x \ dx = = \int cosx(cos^2x) \ dx= \int cosx(1-sin^2x) \ dx# and that's pretty much it

because

# \int cosx(1-sin^2x) \ dx#

# = \int cosx- cosx sin^2x \ dx#

# = sin x- 1/3 sin^3x + C#

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Answer 2

William Abdalla

To find the antiderivative of ( \cos^3(x) ), we can use the reduction formula:

[ \int \cos^n(x) , dx = \frac{\cos^{n-1}(x) \sin(x)}{n} + \frac{n-1}{n} \int \cos^{n-2}(x) , dx ]

For ( \cos^3(x) ), we have ( n = 3 ). Applying the reduction formula:

[ \int \cos^3(x) , dx = \frac{\cos^2(x) \sin(x)}{3} + \frac{2}{3} \int \cos(x) , dx ]

The integral of ( \cos(x) ) is ( \sin(x) ), so:

[ \int \cos^3(x) , dx = \frac{\cos^2(x) \sin(x)}{3} + \frac{2}{3} \sin(x) + C ]

Where ( C ) is the constant of integration. So, the antiderivative of ( \cos^3(x) ) is:

[ \frac{\cos^2(x) \sin(x)}{3} + \frac{2}{3} \sin(x) + C ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.