The integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x\left( an ^5 x+\cot ^5 x ight)}$ is equal to

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The integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x\left(	an ^5 x+\cot ^5 xight)}$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} 	ext {Let } I & =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x\left(	an ^5 x+\cot ^5 xight)} \ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\mathrm{~d} x}{2 \sin x \cos x\left(	an ^5 x+\frac{1}{	an ^5 x}ight)} \ & =\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\sec ^2 x}{	an x\left(\frac{	an ^{10} x+1}{	an ^5 x}ight)} \mathrm{d} x \ & =\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{	an ^4 x \sec ^2 x}{	an ^{10} x+1} \mathrm{~d} x \end{aligned}$ Put $	an ^5 x=\mathrm{t} \Rightarrow 5 	an ^4 x \sec ^2 x \mathrm{~d} x=\mathrm{dt}$ $\begin{aligned} 	herefore \quad \mathrm{I} & =\frac{1}{2} \int_{\frac{1}{9 \sqrt{3}}}^1 \frac{\frac{\mathrm{dt}}{5}}{\mathrm{t}^2+1} \ & =\frac{1}{10}\left[	an ^{-1} \mathrm{t}ight]_{\frac{1}{9 \sqrt{3}}}^1 \ & =\frac{1}{10}\left[	an ^{-1} 1-	an ^{-1}\left(\frac{1}{9 \sqrt{3}}ight)ight] \ & =\frac{1}{10}\left[\frac{\pi}{4}-	an ^{-1}\left(\frac{1}{9 \sqrt{3}}ight)ight]\end{aligned}$

The integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{d x}{\sin 2 x\left(\tan ^5 x+\cot ^5 x\right)}$ is…

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