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

Regarding a non-parts method

Extending that

And that

It's simple to solve the first integral.

The second is simple if you keep that in mind.

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To integrate ( \int (2 + \sin(\frac{x}{2}))^2 \cos(\frac{x}{2}) , dx ) using integration by parts, let's denote:

[ u = (2 + \sin(\frac{x}{2}))^2 ] [ dv = \cos(\frac{x}{2}) , dx ]

Now, we differentiate ( u ) and integrate ( dv ) to find ( du ) and ( v ), respectively:

[ du = 2(2 + \sin(\frac{x}{2})) \cos(\frac{x}{2}) , dx ] [ v = 2\sin(\frac{x}{2}) ]

Now, we apply the integration by parts formula:

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

Substitute the values:

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

[ = (2 + \sin(\frac{x}{2}))^2 \cdot 2\sin(\frac{x}{2}) - 4 \int (2 + \sin(\frac{x}{2})) \sin(\frac{x}{2}) \cos(\frac{x}{2}) , dx ]

Now, the integral on the right side can be solved using a trigonometric identity or by recognizing it as a double angle identity:

[ \sin(\frac{x}{2}) \cos(\frac{x}{2}) = \frac{1}{2} \sin(x) ]

Substituting this into the integral:

[ = (2 + \sin(\frac{x}{2}))^2 \cdot 2\sin(\frac{x}{2}) - 4 \int (2 + \sin(\frac{x}{2})) \frac{1}{2} \sin(x) , dx ]

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

At this point, you can continue integrating by parts or employ other techniques to solve the remaining integral.

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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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