int_0^{frac{pi}{4}} sec^4 x , dx = | vedclass

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(C) We want to evaluate the integral $I = \int_0^{\frac{\pi}{4}} \sec^4 x \, dx$. Using the trigonometric identity $\sec^2 x = 1 + 	an^2 x$,we can re

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$\frac{4}{3}$

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(C) We want to evaluate the integral $I = \int_0^{\frac{\pi}{4}} \sec^4 x \, dx$. Using the trigonometric identity $\sec^2 x = 1 + 	an^2 x$,we can rewrite the integral as: $I = \int_0^{\frac{\pi}{4}} \sec^2 x \cdot \sec^2 x \, dx = \int_0^{\frac{\pi}{4}} (1 + 	an^2 x) \sec^2 x \, dx$. Let $u = 	an x$. Then $du = \sec^2 x \, dx$. When $x = 0$,$u = 	an(0) = 0$. When $x = \frac{\pi}{4}$,$u = 	an(\frac{\pi}{4}) = 1$. Substituting these into the integral: $I = \int_0^1 (1 + u^2) \, du$. Integrating with respect to $u$: $I = [u + \frac{u^3}{3}]_0^1$. Evaluating at the limits: $I = (1 + \frac{1^3}{3}) - (0 + \frac{0^3}{3}) = 1 + \frac{1}{3} = \frac{4}{3}$.

$\int_0^{\frac{\pi}{4}} \sec^4 x \, dx =$

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