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

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(C) Let $I = \int_{\pi/6}^{\pi/3} \frac{4 - \csc^2 x}{\cos^4 x} dx = \int_{\pi/6}^{\pi/3} (4 \sec^4 x - \csc^2 x \sec^4 x) dx$.Using $\sec^2 x = 1 + \

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

Vedclass · Answer

(C) Let $I = \int_{\pi/6}^{\pi/3} \frac{4 - \csc^2 x}{\cos^4 x} dx = \int_{\pi/6}^{\pi/3} (4 \sec^4 x - \csc^2 x \sec^4 x) dx$.Using $\sec^2 x = 1 + 	an^2 x$ and $\csc^2 x = \frac{1}{\sin^2 x}$, we have:$I = \int_{\pi/6}^{\pi/3} (4 \sec^2 x (1 + 	an^2 x) - \frac{1}{\sin^2 x \cos^4 x}) dx$.Substitute $u = 	an x$, then $du = \sec^2 x dx$. When $x = \pi/6, u = 1/\sqrt{3}$; when $x = \pi/3, u = \sqrt{3}$.Note that $\frac{1}{\sin^2 x \cos^4 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^4 x} = \frac{1}{\cos^4 x} + \frac{1}{\sin^2 x \cos^2 x} = \sec^4 x + \sec^2 x \csc^2 x = (1+u^2)^2 + (1+u^2)(1 + 1/u^2) = (1+u^2)^2 + (1+u^2 + 1/u^2 + 1) = (1+u^2)^2 + u^2 + 2 + 1/u^2$.Alternatively, $I = \int_{1/\sqrt{3}}^{\sqrt{3}} (4(1+u^2) - (1+u^2)(1 + 1/u^2)) du = \int_{1/\sqrt{3}}^{\sqrt{3}} (4 + 4u^2 - (1 + 1/u^2 + u^2 + 1)) du = \int_{1/\sqrt{3}}^{\sqrt{3}} (2 + 3u^2 - u^{-2}) du$.$I = [2u + u^3 + \frac{1}{u}]_{1/\sqrt{3}}^{\sqrt{3}} = (2\sqrt{3} + 3\sqrt{3} + \frac{1}{\sqrt{3}}) - (\frac{2}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \sqrt{3}) = (5\sqrt{3} + \frac{1}{\sqrt{3}}) - (\sqrt{3} + \frac{7}{3\sqrt{3}}) = 4\sqrt{3} + \frac{3-7}{3\sqrt{3}} = 4\sqrt{3} - \frac{4}{3\sqrt{3}} = \frac{36-4}{3\sqrt{3}} = \frac{32}{3\sqrt{3}}$.

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

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