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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x(	an ^5 x+\cot ^5 x)}$Using $\sin 2x = 2 \sin x \cos x$ and $\cot x = \frac{1}{	an x}$,we have:$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{2 \sin x \cos x(	an ^5 x+\frac{1}{	an ^5 x})}$Multiply numerator and denominator by $\sec^2 x$:$I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\sec^2 x}{	an x(\frac{	an^{10} x+1}{	an^5 x})} dx = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{	an^4 x \sec^2 x}{	an^{10} x+1} dx$Let $t = 	an^5 x$,then $dt = 5 	an^4 x \sec^2 x dx$,so $	an^4 x \sec^2 x dx = \frac{dt}{5}$.When $x = \frac{\pi}{6}$,$t = (\frac{1}{\sqrt{3}})^5 = \frac{1}{9 \sqrt{3}}$. When $x = \frac{\pi}{4}$,$t = 1^5 = 1$.$I = \frac{1}{2} \int_{\frac{1}{9 \sqrt{3}}}^{1} \frac{dt/5}{t^2+1} = \frac{1}{10} [	an^{-1} t]_{\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 integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x(\tan ^5 x+\cot ^5 x)}$ is equal to

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