The value of int_{frac{pi}{6}}^{frac{pi}{4}} | vedclass

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(C) $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\sin 2 x(	an ^5 x+\cot ^5 x)} dx$Using $\sin 2x = \frac{2 	an x}{1+	an ^2 x}$ and $\cot x =

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$\frac{1}{10}(\frac{\pi}{4}-	an ^{-1}(\frac{1}{9 \sqrt{3}}))$

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(C) $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\sin 2 x(	an ^5 x+\cot ^5 x)} dx$Using $\sin 2x = \frac{2 	an x}{1+	an ^2 x}$ and $\cot x = \frac{1}{	an x}$,we get:$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1+	an ^2 x}{2 	an x(	an ^5 x + \frac{1}{	an ^5 x})} dx$$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\sec ^2 x}{2 	an x(\frac{	an ^{10} x + 1}{	an ^5 x})} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{	an ^4 x \sec ^2 x}{2(	an ^{10} x + 1)} dx$Let $	an x = t$,then $\sec ^2 x dx = dt$.When $x = \frac{\pi}{6}, t = \frac{1}{\sqrt{3}}$. When $x = \frac{\pi}{4}, t = 1$.$I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{t^4}{2(t^{10} + 1)} dt$Let $t^5 = u$,then $5t^4 dt = du$,so $t^4 dt = \frac{du}{5}$.When $t = \frac{1}{\sqrt{3}}, u = (\frac{1}{\sqrt{3}})^5 = \frac{1}{9 \sqrt{3}}$. When $t = 1, u = 1$.$I = \frac{1}{10} \int_{\frac{1}{9 \sqrt{3}}}^{1} \frac{du}{u^2 + 1} = \frac{1}{10} [	an ^{-1} u]_{\frac{1}{9 \sqrt{3}}}^{1}$$I = \frac{1}{10} (	an ^{-1} 1 - 	an ^{-1}(\frac{1}{9 \sqrt{3}})) = \frac{1}{10} (\frac{\pi}{4} - 	an ^{-1}(\frac{1}{9 \sqrt{3}}))$

The value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\sin 2 x(\tan ^5 x+\cot ^5 x)} dx$ is

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