int frac{1}{cos x(1 + cos x)} , dx = | vedclass

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(D) To solve the integral $I = \int \frac{1}{\cos x(1 + \cos x)} \, dx$,we use partial fractions or algebraic manipulation.We can write the integrand

Vedclass · Accepted Answer

$\log |\sec x + 	an x| - 	an \frac{x}{2} + c$

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(D) To solve the integral $I = \int \frac{1}{\cos x(1 + \cos x)} \, dx$,we use partial fractions or algebraic manipulation.We can write the integrand as: $\frac{1}{\cos x(1 + \cos x)} = \frac{(1 + \cos x) - \cos x}{\cos x(1 + \cos x)} = \frac{1}{\cos x} - \frac{1}{1 + \cos x}$.Now,integrate each term separately:$I = \int \sec x \, dx - \int \frac{1}{1 + \cos x} \, dx$.Using the identity $1 + \cos x = 2 \cos^2 \frac{x}{2}$,we get:$I = \log |\sec x + 	an x| - \int \frac{1}{2 \cos^2 \frac{x}{2}} \, dx$.$I = \log |\sec x + 	an x| - \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx$.Integrating $\sec^2 \frac{x}{2}$ gives $2 	an \frac{x}{2}$:$I = \log |\sec x + 	an x| - \frac{1}{2} \cdot 2 	an \frac{x}{2} + c$.$I = \log |\sec x + 	an x| - 	an \frac{x}{2} + c$.

$\int \frac{1}{\cos x(1 + \cos x)} \, dx = $

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