# How do you find an antiderivative of #sin 3x cos^5 3x#?

Try this:

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To find the antiderivative of ( \sin(3x) \cos^5(3x) ), you can use the technique of integration by parts.

Integration by parts formula: [ \int u , dv = uv - \int v , du ]

Let's choose: [ u = \cos^5(3x) \implies du = -5\cos^4(3x) \sin(3x) , dx ] [ dv = \sin(3x) , dx \implies v = -\frac{1}{3} \cos(3x) ]

Now, apply the integration by parts formula:

[ \int \sin(3x) \cos^5(3x) , dx = -\frac{1}{3} \cos^5(3x) \cos(3x) + \frac{5}{3} \int \cos^4(3x) \sin(3x) , dx ]

For the remaining integral, we can use another integration by parts:

Let's choose: [ u = \cos^4(3x) \implies du = -4\cos^3(3x) \sin(3x) , dx ] [ dv = \sin(3x) , dx \implies v = -\frac{1}{3} \cos(3x) ]

Applying the integration by parts formula again:

[ \int \cos^4(3x) \sin(3x) , dx = -\frac{1}{3} \cos^4(3x) \cos(3x) + \frac{4}{3} \int \cos^3(3x) \sin(3x) , dx ]

You'll notice that this new integral involves ( \cos^3(3x) \sin(3x) ), which can be further broken down using the same integration by parts method.

After continuing this process for a few more iterations, you'll eventually arrive at a point where the integrals become solvable using basic trigonometric identities or substitution methods. The process can be quite iterative, but it will lead to the antiderivative of the given function.

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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.

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.

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.

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