# How do you integrate #int sin^2(x) dx# using integration by parts?

where

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To integrate ( \int \sin^2(x) , dx ) using integration by parts, we use the formula:

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

Let's choose:

[ u = \sin(x) ] [ dv = \sin(x) , dx ]

Then, we find:

[ du = \cos(x) , dx ] [ v = -\cos(x) ]

Now, we apply the integration by parts formula:

[ \int \sin^2(x) , dx = -\sin(x) \cos(x) - \int (-\cos(x))(\cos(x) , dx) ]

[ = -\sin(x) \cos(x) + \int \cos^2(x) , dx ]

We can rewrite ( \cos^2(x) ) as ( 1 - \sin^2(x) ):

[ \int \cos^2(x) , dx = \int (1 - \sin^2(x)) , dx ]

[ = \int dx - \int \sin^2(x) , dx ]

[ = x - \int \sin^2(x) , dx ]

Now, let's solve for ( \int \sin^2(x) , dx ):

[ 2\int \sin^2(x) , dx = -\sin(x) \cos(x) + x ]

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

Where ( C ) is the constant of integration.

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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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