The value of int_{pi /4}^{3pi /4} frac{phi}{1 | vedclass

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(A) Let $I = \int_{\pi /4}^{3\pi /4} \frac{\phi}{1 + \sin \phi} \, d\phi$. Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,

Vedclass · Accepted Answer

$\pi 	an \frac{\pi}{8}$

Vedclass · Answer

(A) Let $I = \int_{\pi /4}^{3\pi /4} \frac{\phi}{1 + \sin \phi} \, d\phi$. Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we have: $I = \int_{\pi /4}^{3\pi /4} \frac{\pi - \phi}{1 + \sin(\pi - \phi)} \, d\phi = \int_{\pi /4}^{3\pi /4} \frac{\pi - \phi}{1 + \sin \phi} \, d\phi$. Adding the two expressions for $I$: $2I = \int_{\pi /4}^{3\pi /4} \frac{\phi + \pi - \phi}{1 + \sin \phi} \, d\phi = \pi \int_{\pi /4}^{3\pi /4} \frac{1}{1 + \sin \phi} \, d\phi$. $2I = \pi \int_{\pi /4}^{3\pi /4} \frac{1 - \sin \phi}{\cos^2 \phi} \, d\phi = \pi \int_{\pi /4}^{3\pi /4} (\sec^2 \phi - \sec \phi 	an \phi) \, d\phi$. $2I = \pi [	an \phi - \sec \phi]_{\pi /4}^{3\pi /4}$. $2I = \pi [(	an(3\pi/4) - \sec(3\pi/4)) - (	an(\pi/4) - \sec(\pi/4))]$. $2I = \pi [(-1 - (-\sqrt{2})) - (1 - \sqrt{2})] = \pi [\sqrt{2} - 1 - 1 + \sqrt{2}] = \pi [2\sqrt{2} - 2] = 2\pi(\sqrt{2} - 1)$. $I = \pi(\sqrt{2} - 1)$. Since $	an(\pi/8) = \sqrt{2} - 1$,the value is $\pi 	an(\pi/8)$.

The value of $\int_{\pi /4}^{3\pi /4} \frac{\phi}{1 + \sin \phi} \, d\phi$ is

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