The integral int_{pi /6}^{pi /3} {sec ^{2/3} | vedclass

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Question

(C) Let $I = \int_{\pi/6}^{\pi/3} \sec^{2/3} x \csc^{4/3} x \, dx$.We can rewrite the integrand as:$I = \int_{\pi/6}^{\pi/3} \frac{1}{\cos^{2/3} x \si

Vedclass · Accepted Answer

$3^{7/6} - 3^{5/6}$

Vedclass · Answer

(C) Let $I = \int_{\pi/6}^{\pi/3} \sec^{2/3} x \csc^{4/3} x \, dx$.We can rewrite the integrand as:$I = \int_{\pi/6}^{\pi/3} \frac{1}{\cos^{2/3} x \sin^{4/3} x} \, dx = \int_{\pi/6}^{\pi/3} \frac{1}{\cos^{2/3} x \sin^{2/3} x \cdot \sin^{2/3} x} \, dx$$I = \int_{\pi/6}^{\pi/3} \frac{1}{(\sin x \cos x)^{2/3} \cdot \sin^{2/3} x} \, dx = \int_{\pi/6}^{\pi/3} \frac{1}{	an^{2/3} x \cdot \sin^2 x} \, dx$$I = \int_{\pi/6}^{\pi/3} \frac{\sec^2 x}{	an^{2/3} x} \, dx$.Let $	an x = t$,then $\sec^2 x \, dx = dt$.When $x = \pi/6$,$t = 	an(\pi/6) = 1/\sqrt{3} = 3^{-1/2}$.When $x = \pi/3$,$t = 	an(\pi/3) = \sqrt{3} = 3^{1/2}$.$I = \int_{3^{-1/2}}^{3^{1/2}} t^{-2/3} \, dt = \left[ \frac{t^{1/3}}{1/3} ight]_{3^{-1/2}}^{3^{1/2}} = 3 \left[ t^{1/3} ight]_{3^{-1/2}}^{3^{1/2}}$$I = 3 \left( (3^{1/2})^{1/3} - (3^{-1/2})^{1/3} ight) = 3 \left( 3^{1/6} - 3^{-1/6} ight)$$I = 3^{1 + 1/6} - 3^{1 - 1/6} = 3^{7/6} - 3^{5/6}$.

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

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