The integral int_{frac{pi}{6}}^{frac{pi}{3}} | vedclass

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Question

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \, dx$ $= \int_{\frac{\pi}{6}}^{\frac{\pi}

Vedclass · Accepted Answer

$3^{\frac{7}{6}}-3^{\frac{5}{6}}$

Vedclass · Answer

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \, dx$ $= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\cos^{\frac{2}{3}} x \sin^{\frac{4}{3}} x} \, dx$ $= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\left(\frac{\sin x}{\cos x}ight)^{\frac{4}{3}} \cos^2 x} \, dx$ $= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec^2 x}{	an^{\frac{4}{3}} x} \, dx$ Let $	an x = t$,then $\sec^2 x \, dx = dt$. When $x = \frac{\pi}{6}$,$t = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} = 3^{-\frac{1}{2}}$. When $x = \frac{\pi}{3}$,$t = 	an(\frac{\pi}{3}) = \sqrt{3} = 3^{\frac{1}{2}}$. $I = \int_{3^{-\frac{1}{2}}}^{3^{\frac{1}{2}}} t^{-\frac{4}{3}} \, dt = \left[ \frac{t^{-\frac{1}{3}}}{-\frac{1}{3}} ight]_{3^{-\frac{1}{2}}}^{3^{\frac{1}{2}}} = -3 \left[ t^{-\frac{1}{3}} ight]_{3^{-\frac{1}{2}}}^{3^{\frac{1}{2}}}$ $= -3 \left( (3^{\frac{1}{2}})^{-\frac{1}{3}} - (3^{-\frac{1}{2}})^{-\frac{1}{3}} ight) = -3 \left( 3^{-\frac{1}{6}} - 3^{\frac{1}{6}} ight)$ $= 3 \cdot 3^{\frac{1}{6}} - 3 \cdot 3^{-\frac{1}{6}} = 3^{\frac{7}{6}} - 3^{\frac{5}{6}}$

The integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \, dx$ is equal to

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