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How do you find the antiderivative of #cos^(2)3x dx#?

Answer 1

The answer is #=(sin6x+6x)/12+C#

We use #cos2x=2cos^2x-1# Adapting this #cos6x=2cos^2(3x)-1# Therefore, #Cos^2(3x)=(cos6x+1)/2# So #intCos^2(3x)dx=int((cos6x+1)dx)/2# #=1/2(sin(6x)/6+x)+C# #(sin6x+6x)/12+C#

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Answer 2

William Abdalla

To find the antiderivative of ( \cos^2(3x) , dx ), you can use the trigonometric identity ( \cos^2(\theta) = \frac{1}{2}(1 + \cos(2\theta)) ). Applying this identity, we have ( \cos^2(3x) = \frac{1}{2}(1 + \cos(6x)) ). Now, integrate each term separately.

The antiderivative of ( \frac{1}{2} , dx ) is ( \frac{1}{2}x ).

The antiderivative of ( \cos(6x) , dx ) is ( \frac{1}{6} \sin(6x) ).

Therefore, the antiderivative of ( \cos^2(3x) , dx ) is ( \frac{1}{2}x + \frac{1}{12} \sin(6x) + 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.