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

Answer 1

Ezra Adams

#1/3sin3x-1/9sin^3 3x+c#

#intcos^3 3xdx#

#=intcos3xcos^2 3xdx#

#=intcos3x(1-sin^2 3x)dx#

#I=int(cos3x-cos3xsin^2 3x)dx#

now #d/(dx)(sin^3 3x)=3 xx3xxsin^2 3xcos3x=9sin^2 3xcos3x#

#I=1/3sin3x-1/9sin^3 3x+c#

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Answer 2

Charles Anctil

To find the antiderivative of ( (\cos(3x))^3 ), you can use the substitution method. Let ( u = \sin(3x) ), then ( du = 3\cos(3x)dx ). Rearranging, we get ( \frac{du}{3} = \cos(3x)dx ). Now, substitute ( u ) and ( \frac{du}{3} ) into the integral:

[ \int (\cos(3x))^3 dx = \int u^3 \cdot \frac{du}{3} = \frac{1}{3} \int u^3 du ]

Integrating ( u^3 ) with respect to ( u ), we get:

[ \frac{1}{3} \int u^3 du = \frac{1}{3} \cdot \frac{u^4}{4} + C ]

Substitute back ( u = \sin(3x) ):

[ = \frac{1}{12} \sin^4(3x) + C ]

Therefore, the antiderivative of ( (\cos(3x))^3 ) is ( \frac{1}{12} \sin^4(3x) + C ), where ( C ) is the constant of integration.

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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.