The integral int_{pi /6}^{pi /4} {frac{{dx}}{ | vedclass

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(B) Let $I = \int_{\pi /6}^{\pi /4} \frac{dx}{\sin 2x (	an^5 x + \cot^5 x)}$.Using $\sin 2x = \frac{2	an x}{1+	an^2 x}$,we have $\frac{1}{\sin 2x}

Vedclass · Accepted Answer

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

Vedclass · Answer

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

The integral $\int_{\pi /6}^{\pi /4} {\frac{{dx}}{{\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $ equals

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