For n geq 2,if I_n = int sec^n x , dx,then I_ | vedclass

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(B) Given $I_n = \int \sec^n x \, dx$. First,calculate $I_2 = \int \sec^2 x \, dx = 	an x + c_1$. Next,calculate $I_4 = \int \sec^4 x \, dx = \int \s

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$\frac{1}{3} \sec^2 x 	an x + c$

Vedclass · Answer

(B) Given $I_n = \int \sec^n x \, dx$. First,calculate $I_2 = \int \sec^2 x \, dx = 	an x + c_1$. Next,calculate $I_4 = \int \sec^4 x \, dx = \int \sec^2 x \cdot \sec^2 x \, dx$. Using the identity $\sec^2 x = 1 + 	an^2 x$,we get: $I_4 = \int (1 + 	an^2 x) \sec^2 x \, dx$. Let $u = 	an x$,then $du = \sec^2 x \, dx$. $I_4 = \int (1 + u^2) \, du = u + \frac{u^3}{3} + c_2 = 	an x + \frac{	an^3 x}{3} + c_2$. Now,compute $I_4 - \frac{2}{3} I_2$: $I_4 - \frac{2}{3} I_2 = (	an x + \frac{	an^3 x}{3} + c_2) - \frac{2}{3} (	an x + c_1)$. $= 	an x + \frac{	an^3 x}{3} - \frac{2}{3} 	an x + c$. $= \frac{1}{3} 	an x + \frac{1}{3} 	an^3 x + c$. $= \frac{1}{3} 	an x (1 + 	an^2 x) + c$. Since $1 + 	an^2 x = \sec^2 x$,we have: $= \frac{1}{3} 	an x \sec^2 x + c$.

For $n \geq 2$,if $I_n = \int \sec^n x \, dx$,then $I_4 - \frac{2}{3} I_2 =$

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