int_{-frac{pi}{2}}^{frac{pi}{2}} log left(fra | vedclass

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Question

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin 	heta}{2+\sin 	heta}ight) d 	heta$. Define $f(	heta) = \log \left(\fr

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Vedclass · Answer

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin 	heta}{2+\sin 	heta}ight) d 	heta$. Define $f(	heta) = \log \left(\frac{2-\sin 	heta}{2+\sin 	heta}ight)$. Now,check for parity by calculating $f(-	heta)$: $f(-	heta) = \log \left(\frac{2-\sin(-	heta)}{2+\sin(-	heta)}ight) = \log \left(\frac{2+\sin 	heta}{2-\sin 	heta}ight)$. Using the property $\log(x^{-1}) = -\log(x)$,we get: $f(-	heta) = \log \left(\left(\frac{2-\sin 	heta}{2+\sin 	heta}ight)^{-1}ight) = -\log \left(\frac{2-\sin 	heta}{2+\sin 	heta}ight) = -f(	heta)$. Since $f(-	heta) = -f(	heta)$,the function $f(	heta)$ is an odd function. According to the property of definite integrals,$\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function. Therefore,$I = 0$.

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$ is equal to

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