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

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Aicharya · Accepted Answer

Let $\mathrm{I}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \mathrm{~d} x$ $=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{\cos ^{\frac{2}{3}} x \cdot \sin ^{\frac{4}{3}} x}$ $=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{\frac{\sin ^{\frac{4}{3}} x}{\cos ^{\frac{4}{3}} x} \cdot \cos ^2 x}$ $=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec ^2 x}{	an ^{\frac{4}{3}} x} \mathrm{~d} x$ $\begin{aligned} & 	ext { Put } 	an x=\mathrm{t} \ & \Rightarrow \sec ^2 x \mathrm{~d} x=\mathrm{dt} \ 	herefore \quad \mathrm{I} & =\int_{\frac{1}{\sqrt{3}}} \mathrm{t}^{\frac{\sqrt{3}}{3}} \frac{\mathrm{dt}}{\mathrm{x}^{-1}} \ & =\left[-3 \mathrm{t}^{-\frac{1}{3}}ight]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \ & =-3\left[(\sqrt{3})^{\frac{-1}{3}}-\left(\frac{1}{\sqrt{3}}ight)^{\frac{-1}{3}}ight] \ & =-3\left(3^{-\frac{1}{6}}-3^{\frac{1}{6}}ight) \ & =3^{\frac{7}{6}}-3^{\frac{5}{6}}\end{aligned}$

The integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}}…

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