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

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Question

(A) We know that $1 + \cos x = 2 \cos^2 \frac{x}{2}$.Substituting this into the integral:$\int \frac{1}{\sqrt{2 \cos^2 \frac{x}{2}}} \, dx = \int \fra

Vedclass · Accepted Answer

$\sqrt{2} \log \left| \sec \frac{x}{2} + 	an \frac{x}{2} ight| + K$

Vedclass · Answer

(A) We know that $1 + \cos x = 2 \cos^2 \frac{x}{2}$.Substituting this into the integral:$\int \frac{1}{\sqrt{2 \cos^2 \frac{x}{2}}} \, dx = \int \frac{1}{\sqrt{2} \cos \frac{x}{2}} \, dx$$= \frac{1}{\sqrt{2}} \int \sec \frac{x}{2} \, dx$Using the formula $\int \sec 	heta \, d	heta = \log |\sec 	heta + 	an 	heta| + C$ and applying the chain rule for $\frac{x}{2}$ (where the derivative is $\frac{1}{2}$):$= \frac{1}{\sqrt{2}} \cdot \frac{\log |\sec \frac{x}{2} + 	an \frac{x}{2}|}{1/2} + K$$= \frac{2}{\sqrt{2}} \log |\sec \frac{x}{2} + 	an \frac{x}{2}| + K$$= \sqrt{2} \log |\sec \frac{x}{2} + 	an \frac{x}{2}| + K$.

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

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