# What is the area enclosed by #r=thetacostheta-2sin(theta/2-pi) # for #theta in [pi/4,pi]#?

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To find the area enclosed by the curve ( r = \theta \cos(\theta) - 2 \sin\left(\frac{\theta}{2} - \pi\right) ) for ( \theta ) in ( [\frac{\pi}{4}, \pi] ), you need to integrate ( \frac{1}{2} r^2 ) with respect to ( \theta ) over the given interval. The integral will give you the area of one loop of the curve. Then, you can subtract the area under the curve for ( \theta ) from ( \frac{\pi}{4} ) to ( \frac{\pi}{2} ) from the area under the curve for ( \theta ) from ( \frac{\pi}{2} ) to ( \pi ). This subtraction accounts for the loops in opposite directions.

The expression for the area ( A ) enclosed by the curve is:

[ A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{2} \left(\theta \cos(\theta) - 2 \sin\left(\frac{\theta}{2} - \pi\right)\right)^2 , d\theta - \int_{\frac{\pi}{2}}^{\pi} \frac{1}{2} \left(\theta \cos(\theta) - 2 \sin\left(\frac{\theta}{2} - \pi\right)\right)^2 , d\theta ]

You can then calculate each integral separately using numerical methods or appropriate techniques like integration by parts or substitution.

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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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