int_{-pi / 4}^{pi / 4} cos^{-8} x , dx = | vedclass

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(C) Let $I = \int_{-\pi / 4}^{\pi / 4} \frac{1}{\cos^8 x} \, dx = \int_{-\pi / 4}^{\pi / 4} (\sec^2 x)^3 \sec^2 x \, dx$. Since the integrand is an ev

Vedclass · Accepted Answer

$\frac{192}{35}$

Vedclass · Answer

(C) Let $I = \int_{-\pi / 4}^{\pi / 4} \frac{1}{\cos^8 x} \, dx = \int_{-\pi / 4}^{\pi / 4} (\sec^2 x)^3 \sec^2 x \, dx$. Since the integrand is an even function,$I = 2 \int_{0}^{\pi / 4} (1 + 	an^2 x)^3 \sec^2 x \, dx$. Let $	an x = t$,then $\sec^2 x \, dx = dt$. When $x = 0, t = 0$ and when $x = \pi / 4, t = 1$. $I = 2 \int_{0}^{1} (1 + t^2)^3 \, dt = 2 \int_{0}^{1} (1 + 3t^2 + 3t^4 + t^6) \, dt$. $I = 2 \left[ t + t^3 + \frac{3t^5}{5} + \frac{t^7}{7} ight]_{0}^{1} = 2 \left( 1 + 1 + \frac{3}{5} + \frac{1}{7} ight)$. $I = 2 \left( 2 + \frac{21 + 5}{35} ight) = 2 \left( 2 + \frac{26}{35} ight) = 2 \left( \frac{70 + 26}{35} ight) = 2 \left( \frac{96}{35} ight) = \frac{192}{35}$.

$\int_{-\pi / 4}^{\pi / 4} \cos^{-8} x \, dx =$

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