Evaluate the integral: int sec^5 x , dx | vedclass

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Question

(B) Let $I = \int \sec^5 x \, dx = \int \sec^3 x \cdot \sec^2 x \, dx$. Using integration by parts, let $u = \sec^3 x$ and $dv = \sec^2 x \, dx$. Then

Vedclass · Accepted Answer

$\frac{1}{4} 	an^3 x \sec x + \frac{5}{8} \sec x 	an x + \frac{3}{8} \ln |\sec x + 	an x| + c$

Vedclass · Answer

(B) Let $I = \int \sec^5 x \, dx = \int \sec^3 x \cdot \sec^2 x \, dx$. Using integration by parts, let $u = \sec^3 x$ and $dv = \sec^2 x \, dx$. Then $du = 3 \sec^2 x (\sec x 	an x) \, dx = 3 \sec^3 x 	an x \, dx$ and $v = 	an x$. $I = \sec^3 x 	an x - \int 3 \sec^3 x 	an^2 x \, dx$. Since $	an^2 x = \sec^2 x - 1$, we have: $I = \sec^3 x 	an x - 3 \int \sec^3 x (\sec^2 x - 1) \, dx$. $I = \sec^3 x 	an x - 3 \int \sec^5 x \, dx + 3 \int \sec^3 x \, dx$. $4I = \sec^3 x 	an x + 3 \int \sec^3 x \, dx$. Using the standard formula $\int \sec^3 x \, dx = \frac{1}{2} (\sec x 	an x + \ln |\sec x + 	an x|) + c_1$: $4I = \sec^3 x 	an x + \frac{3}{2} \sec x 	an x + \frac{3}{2} \ln |\sec x + 	an x| + C$. $I = \frac{1}{4} \sec^3 x 	an x + \frac{3}{8} \sec x 	an x + \frac{3}{8} \ln |\sec x + 	an x| + c$. Since $\sec^3 x = \sec x (1 + 	an^2 x) = \sec x + \sec x 	an^2 x$, the expression can also be written as: $I = \frac{1}{4} \sec x 	an^3 x + \frac{5}{8} \sec x 	an x + \frac{3}{8} \ln |\sec x + 	an x| + c$.

Evaluate the integral: $\int \sec^5 x \, dx$

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