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$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x=$

$\int_{-4}^{4} \log \left(\frac{8-x}{8+x}\right) d x=$

The value of $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \,dx$ is equal to

$\int_0^{\frac{\pi}{4}} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x=$

By using the properties of definite integrals,evaluate the integral $\int_{0}^{4}|x-1| d x$.

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