The value of the integral int_{pi/6}^{pi/3} f | vedclass

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(C) Let $I = \int_{\pi/6}^{\pi/3} \frac{4 - \csc^2 x}{\cos^4 x} dx$.We can rewrite the integrand as:$\frac{4}{\cos^4 x} - \frac{\csc^2 x}{\cos^4 x} =

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$\frac{32}{3\sqrt{3}}$

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(C) Let $I = \int_{\pi/6}^{\pi/3} \frac{4 - \csc^2 x}{\cos^4 x} dx$.We can rewrite the integrand as:$\frac{4}{\cos^4 x} - \frac{\csc^2 x}{\cos^4 x} = 4\sec^4 x - \frac{1}{\sin^2 x \cos^4 x}$.Using the identity $1 = \sin^2 x + \cos^2 x$,we have:$\frac{1}{\sin^2 x \cos^4 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^4 x} = \frac{1}{\cos^4 x} + \frac{1}{\sin^2 x \cos^2 x} = \sec^4 x + \frac{4}{\sin^2(2x)} = \sec^4 x + 4\csc^2(2x)$.Substituting this back into the integral:$I = \int_{\pi/6}^{\pi/3} (4\sec^4 x - (\sec^4 x + 4\csc^2(2x))) dx = \int_{\pi/6}^{\pi/3} (3\sec^4 x - 4\csc^2(2x)) dx$.Since $\sec^4 x = (1 + 	an^2 x)\sec^2 x$,we have:$I = \int_{\pi/6}^{\pi/3} (3(1 + 	an^2 x)\sec^2 x - 4\csc^2(2x)) dx$.Let $u = 	an x$,then $du = \sec^2 x dx$. When $x = \pi/6, u = 1/\sqrt{3}$. When $x = \pi/3, u = \sqrt{3}$.$I = [3(u + \frac{u^3}{3}) + 2\cot(2x)]_{\pi/6}^{\pi/3} = [3u + u^3 + 2\cot(2x)]_{\pi/6}^{\pi/3}$.Evaluating at the limits:At $x = \pi/3: 3(\sqrt{3}) + (\sqrt{3})^3 + 2\cot(2\pi/3) = 3\sqrt{3} + 3\sqrt{3} + 2(-1/\sqrt{3}) = 6\sqrt{3} - 2/\sqrt{3} = \frac{18-2}{\sqrt{3}} = \frac{16}{\sqrt{3}}$.At $x = \pi/6: 3(1/\sqrt{3}) + (1/\sqrt{3})^3 + 2\cot(\pi/3) = \sqrt{3} + 1/(3\sqrt{3}) + 2(1/\sqrt{3}) = \sqrt{3} + 1/(3\sqrt{3}) + 2/\sqrt{3} = \sqrt{3} + 7/(3\sqrt{3}) = \frac{9+7}{3\sqrt{3}} = \frac{16}{3\sqrt{3}}$.$I = \frac{16}{\sqrt{3}} - \frac{16}{3\sqrt{3}} = \frac{48-16}{3\sqrt{3}} = \frac{32}{3\sqrt{3}}$.

The value of the integral $\int_{\pi/6}^{\pi/3} \frac{4 - \csc^2 x}{\cos^4 x} dx$ is:

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