int frac{1-cos x}{cos x(1+cos x)} d x= | vedclass

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(B) We have the integral $I = \int \frac{1-\cos x}{\cos x(1+\cos x)} d x$. Rewrite the numerator as $(1+\cos x) - 2\cos x$: $I = \int \frac{1+\cos x -

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$\log |\sec x+	an x|-2(\operatorname{cosec} x-\cot x)+C$

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(B) We have the integral $I = \int \frac{1-\cos x}{\cos x(1+\cos x)} d x$. Rewrite the numerator as $(1+\cos x) - 2\cos x$: $I = \int \frac{1+\cos x - 2\cos x}{\cos x(1+\cos x)} d x = \int \frac{1}{\cos x} d x - 2 \int \frac{1}{1+\cos x} d x$. Using the identity $\int \sec x d x = \log |\sec x + 	an x| + C$: $I = \log |\sec x + 	an x| - 2 \int \frac{1}{2\cos^2(x/2)} d x = \log |\sec x + 	an x| - \int \sec^2(x/2) d x$. Alternatively,using the rationalization method: $I = \int \sec x d x - 2 \int \frac{1-\cos x}{1-\cos^2 x} d x = \int \sec x d x - 2 \int \frac{1-\cos x}{\sin^2 x} d x$. $I = \int \sec x d x - 2 \int (\operatorname{cosec}^2 x - \operatorname{cosec} x \cot x) d x$. Integrating term by term: $I = \log |\sec x + 	an x| - 2(-\cot x - (-\operatorname{cosec} x)) + C$. $I = \log |\sec x + 	an x| - 2(\operatorname{cosec} x - \cot x) + C$.

$\int \frac{1-\cos x}{\cos x(1+\cos x)} d x=$

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