The value of int sec^3 x , dx is | vedclass

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Question

(A) Let $I = \int \sec^3 x \, dx = \int \sec x \cdot \sec^2 x \, dx$. Using integration by parts,taking $u = \sec x$ and $dv = \sec^2 x \, dx$,we have

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$\frac{1}{2} [\sec x 	an x + \log |\sec x + 	an x|] + C$

Vedclass · Answer

(A) Let $I = \int \sec^3 x \, dx = \int \sec x \cdot \sec^2 x \, dx$. Using integration by parts,taking $u = \sec x$ and $dv = \sec^2 x \, dx$,we have $du = \sec x 	an x \, dx$ and $v = 	an x$. $I = \sec x 	an x - \int 	an x (\sec x 	an x) \, dx$ $I = \sec x 	an x - \int \sec x 	an^2 x \, dx$ $I = \sec x 	an x - \int \sec x (\sec^2 x - 1) \, dx$ $I = \sec x 	an x - \int \sec^3 x \, dx + \int \sec x \, dx$ $I = \sec x 	an x - I + \log |\sec x + 	an x|$ $2I = \sec x 	an x + \log |\sec x + 	an x|$ $I = \frac{1}{2} [\sec x 	an x + \log |\sec x + 	an x|] + C$.

The value of $\int \sec^3 x \, dx$ is

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