The value of int sqrt{1+sec x} , dx is | vedclass

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(B) Let $I = \int \sqrt{1+\sec x} \, dx$. We know that $1+\sec x = 1 + \frac{1}{\cos x} = \frac{\cos x + 1}{\cos x} = \frac{2 \cos^2 \frac{x}{2}}{\cos

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$2 \sin ^{-1}(\sqrt{2} \sin \frac{x}{2})+C$

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(B) Let $I = \int \sqrt{1+\sec x} \, dx$. We know that $1+\sec x = 1 + \frac{1}{\cos x} = \frac{\cos x + 1}{\cos x} = \frac{2 \cos^2 \frac{x}{2}}{\cos x}$. So,$I = \int \frac{\sqrt{2} \cos \frac{x}{2}}{\sqrt{\cos x}} \, dx$. Using $\cos x = 1 - 2 \sin^2 \frac{x}{2}$,we get $I = \int \frac{\sqrt{2} \cos \frac{x}{2}}{\sqrt{1 - 2 \sin^2 \frac{x}{2}}} \, dx$. Let $t = \sqrt{2} \sin \frac{x}{2}$. Then $dt = \sqrt{2} \cdot \frac{1}{2} \cos \frac{x}{2} \, dx = \frac{1}{\sqrt{2}} \cos \frac{x}{2} \, dx$. This implies $\sqrt{2} \cos \frac{x}{2} \, dx = 2 \, dt$. Substituting these into the integral,we get $I = \int \frac{2 \, dt}{\sqrt{1 - t^2}} = 2 \sin^{-1}(t) + C$. Substituting back $t = \sqrt{2} \sin \frac{x}{2}$,we obtain $I = 2 \sin^{-1}(\sqrt{2} \sin \frac{x}{2}) + C$.

The value of $\int \sqrt{1+\sec x} \, dx$ is

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