# How do you evaluate #int sinx/(1+cos^2x)# from #[pi/2, pi]#?

So:

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To evaluate ( \int_{\frac{\pi}{2}}^{\pi} \frac{\sin(x)}{1 + \cos^2(x)} , dx ), we can use the substitution method. Let ( u = \cos(x) ), then ( du = -\sin(x) , dx ).

When ( x = \frac{\pi}{2} ), ( u = \cos\left(\frac{\pi}{2}\right) = 0 ), and when ( x = \pi ), ( u = \cos(\pi) = -1 ).

So, the integral becomes:

[ -\int_{0}^{-1} \frac{1}{1 + u^2} , du ]

This integral can be evaluated using the arctangent function.

[ -\int_{0}^{-1} \frac{1}{1 + u^2} , du = -\left[\arctan(u)\right]_{0}^{-1} ]

[ = -\left[\arctan(-1) - \arctan(0)\right] ]

[ = -\left[-\frac{\pi}{4} - 0\right] ]

[ = \frac{\pi}{4} ]

So, ( \int_{\frac{\pi}{2}}^{\pi} \frac{\sin(x)}{1 + \cos^2(x)} , dx = \frac{\pi}{4} ).

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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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