# How do you integrate #int x^2cos(3x)# by integration by parts method?

Integration by parts formula

we now have to use IBP on the integral highlighted in blue

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To integrate ( \int x^2 \cos(3x) ) using integration by parts:

Let ( u = x^2 ) and ( dv = \cos(3x) , dx ).

Then, ( du = 2x , dx ) and ( v = \frac{1}{3} \sin(3x) ).

Applying the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

We have:

[ \int x^2 \cos(3x) , dx = \frac{x^2}{3} \sin(3x) - \int \frac{1}{3} \sin(3x) \cdot 2x , dx ]

Integrate the remaining integral, ( \int \frac{1}{3} \sin(3x) \cdot 2x , dx ), by parts again if necessary.

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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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