int_{-pi / 4}^{pi / 4} frac{dx}{1+cos 2x} i | vedclass

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(A) Given the integral $ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{1+\cos 2x} $.Using the trigonometric identity $ 1+\cos 2x = 2\cos^2 x $,w

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$ 1 $

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(A) Given the integral $ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{1+\cos 2x} $.Using the trigonometric identity $ 1+\cos 2x = 2\cos^2 x $,we have:$ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{2\cos^2 x} = \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x \, dx $.Since $ f(x) = \sec^2 x $ is an even function,we can use the property $ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx $.Thus,$ I = \frac{1}{2} 	imes 2 \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx = \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx $.Evaluating the integral,we get $ [	an x]_{0}^{\frac{\pi}{4}} = 	an(\frac{\pi}{4}) - 	an(0) = 1 - 0 = 1 $.

$ \int_{-\pi / 4}^{\pi / 4} \frac{dx}{1+\cos 2x} $ is equal to

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