int e^{-2x} left( frac{1 - sin 2x}{1 + cos 2x | vedclass

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Question

(A) We have the integral $I = \int e^{-2x} \left( \frac{1 - \sin 2x}{1 + \cos 2x} \right) dx$. Using trigonometric identities,$1 - \sin 2x = 1 - 2 \si

Vedclass · Accepted Answer

$\frac{1}{2} e^{-2x} 	an x + C$

Vedclass · Answer

(A) We have the integral $I = \int e^{-2x} \left( \frac{1 - \sin 2x}{1 + \cos 2x} ight) dx$. Using trigonometric identities,$1 - \sin 2x = 1 - 2 \sin x \cos x$ and $1 + \cos 2x = 2 \cos^2 x$. So,the expression becomes $\frac{1 - 2 \sin x \cos x}{2 \cos^2 x} = \frac{1}{2} \sec^2 x - 	an x$. Thus,$I = \int e^{-2x} (\frac{1}{2} \sec^2 x - 	an x) dx = \frac{1}{2} \int e^{-2x} \sec^2 x dx - \int e^{-2x} 	an x dx$. Applying integration by parts to the first term: $\int e^{-2x} \sec^2 x dx = e^{-2x} 	an x - \int (-2e^{-2x}) 	an x dx = e^{-2x} 	an x + 2 \int e^{-2x} 	an x dx$. Substituting this back into $I$: $I = \frac{1}{2} [e^{-2x} 	an x + 2 \int e^{-2x} 	an x dx] - \int e^{-2x} 	an x dx$. $I = \frac{1}{2} e^{-2x} 	an x + \int e^{-2x} 	an x dx - \int e^{-2x} 	an x dx = \frac{1}{2} e^{-2x} 	an x + C$.

$\int e^{-2x} \left( \frac{1 - \sin 2x}{1 + \cos 2x} \right) dx = $

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