int cos 2theta log left( frac{cos theta + sin | vedclass

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(D) Let $I = \int \cos 2	heta \log \left( \frac{\cos 	heta + \sin 	heta }{\cos 	heta - \sin 	heta } ight) d	heta$. First,simplify the logarith

Vedclass · Accepted Answer

$\frac{1}{2} \sin 2	heta \log 	an \left( \frac{\pi }{4} + 	heta ight) - \frac{1}{2} \log \sec 2	heta $

Vedclass · Answer

(D) Let $I = \int \cos 2	heta \log \left( \frac{\cos 	heta + \sin 	heta }{\cos 	heta - \sin 	heta } ight) d	heta$. First,simplify the logarithmic term: $\log \left( \frac{\cos 	heta + \sin 	heta }{\cos 	heta - \sin 	heta } ight) = \log \left( \frac{1 + 	an 	heta }{1 - 	an 	heta } ight) = \log 	an \left( \frac{\pi }{4} + 	heta ight)$. Let $u = \log 	an \left( \frac{\pi }{4} + 	heta ight)$ and $dv = \cos 2	heta d	heta$. Then $du = \frac{1}{	an(\frac{\pi}{4} + 	heta)} \cdot \sec^2(\frac{\pi}{4} + 	heta) d	heta = \frac{\cos(\frac{\pi}{4} + 	heta)}{\sin(\frac{\pi}{4} + 	heta)} \cdot \frac{1}{\cos^2(\frac{\pi}{4} + 	heta)} d	heta = \frac{1}{\sin(\frac{\pi}{4} + 	heta)\cos(\frac{\pi}{4} + 	heta)} d	heta = \frac{2}{\sin(\frac{\pi}{2} + 2	heta)} d	heta = 2 \sec 2	heta d	heta$. Also,$v = \int \cos 2	heta d	heta = \frac{1}{2} \sin 2	heta$. Using integration by parts $\int u dv = uv - \int v du$: $I = \frac{1}{2} \sin 2	heta \log 	an \left( \frac{\pi }{4} + 	heta ight) - \int \left( \frac{1}{2} \sin 2	heta ight) (2 \sec 2	heta) d	heta$. $I = \frac{1}{2} \sin 2	heta \log 	an \left( \frac{\pi }{4} + 	heta ight) - \int 	an 2	heta d	heta$. $I = \frac{1}{2} \sin 2	heta \log 	an \left( \frac{\pi }{4} + 	heta ight) - \frac{1}{2} \log |\sec 2	heta| + C$.

$\int \cos 2\theta \log \left( \frac{\cos \theta + \sin \theta }{\cos \theta - \sin \theta } \right) d\theta = $

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