int_{pi / 6}^{pi / 3} cos^{-4} x , dx = | vedclass

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(D) We need to evaluate the integral $I = \int_{\pi / 6}^{\pi / 3} \cos^{-4} x \, dx = \int_{\pi / 6}^{\pi / 3} \sec^4 x \, dx$. Using the identity $\

Vedclass · Accepted Answer

$\frac{44}{9 \sqrt{3}}$

Vedclass · Answer

(D) We need to evaluate the integral $I = \int_{\pi / 6}^{\pi / 3} \cos^{-4} x \, dx = \int_{\pi / 6}^{\pi / 3} \sec^4 x \, dx$. Using the identity $\sec^2 x = 1 + 	an^2 x$,we can write $\sec^4 x = \sec^2 x \cdot \sec^2 x = (1 + 	an^2 x) \sec^2 x$. So,$I = \int_{\pi / 6}^{\pi / 3} (1 + 	an^2 x) \sec^2 x \, dx$. Let $u = 	an x$,then $du = \sec^2 x \, dx$. When $x = \pi / 6$,$u = 	an(\pi / 6) = 1 / \sqrt{3}$. When $x = \pi / 3$,$u = 	an(\pi / 3) = \sqrt{3}$. Substituting these into the integral: $I = \int_{1 / \sqrt{3}}^{\sqrt{3}} (1 + u^2) \, du = [u + \frac{u^3}{3}]_{1 / \sqrt{3}}^{\sqrt{3}}$. Evaluating at the limits: $I = (\sqrt{3} + \frac{(\sqrt{3})^3}{3}) - (\frac{1}{\sqrt{3}} + \frac{(1 / \sqrt{3})^3}{3}) = (\sqrt{3} + \sqrt{3}) - (\frac{1}{\sqrt{3}} + \frac{1}{9 \sqrt{3}}) = 2 \sqrt{3} - (\frac{9 + 1}{9 \sqrt{3}}) = 2 \sqrt{3} - \frac{10}{9 \sqrt{3}}$. To combine,$I = \frac{2 \sqrt{3} \cdot 9 \sqrt{3} - 10}{9 \sqrt{3}} = \frac{2 \cdot 9 \cdot 3 - 10}{9 \sqrt{3}} = \frac{54 - 10}{9 \sqrt{3}} = \frac{44}{9 \sqrt{3}}$. Thus,the correct option is $D$.

$\int_{\pi / 6}^{\pi / 3} \cos^{-4} x \, dx =$

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