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

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

Aicharya · Accepted Answer

$\begin{aligned} & I=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\sin 2 x\left(	an ^5 x+\cot ^5 xight)} d x \ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\frac{2 	an x}{1+	an ^2 x}\left(	an ^5 x+\frac{1}{	an ^5 x}ight)} d x \ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\left(1+	an ^2 xight)}{2 	an x\left(	an ^5 x+\frac{1}{	an ^5 x}ight)} d x \end{aligned}$ $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\sec ^2 x}{2 	an x\left(	an ^5 x+\frac{1}{	an ^5 x}ight)} d x$ Let $	an x=\mathrm{t}$ $\sec ^2 x \mathrm{~d} x=\mathrm{dt}$ when $x=\frac{\pi}{6}, \mathrm{t}=\frac{1}{\sqrt{3}}$ when $x=\frac{\pi}{4}, t=1$ $\begin{array}{ll}	herefore \quad & I=\int_{\frac{1}{\sqrt{3}}}^1 \frac{d t}{2 t\left(t^5+\frac{1}{t^5}ight)} \ & I=\int_{\frac{1}{\sqrt{3}}}^1 \frac{t^4}{2\left(t^{10}+1ight)} d t \ & 	ext { Let } t^5=u \ 	herefore \quad & 5 t^4 d t=d u\end{array}$ When $\mathrm{t}=\frac{1}{\sqrt{3}}, \mathrm{u}=3^{\frac{-5}{2}}$ When $\mathrm{t}=1, \mathrm{u}=1$ $\begin{aligned} 	herefore \quad I & =\frac{1}{10} \int_{\frac{-5}{32}}^1 \frac{d u}{\left(u^2+1ight)} \ & =\frac{1}{10}\left(	an ^{-1} uight)_{\frac{-5}{2}}^1 \ & =\frac{1}{10}\left[	an ^{-1} 1-	an ^{-1}\left(3^{\frac{-5}{2}}ight)ight] \ & =\frac{1}{10}\left[\frac{\pi}{4}-	an ^{-1} \frac{1}{\left(3^{\frac{5}{2}}ight)}ight] \ 	herefore \quad I & =\frac{1}{10}\left[\frac{\pi}{4}-	an ^{-1}\left(\frac{1}{9 \sqrt{3}}ight)ight] \end{aligned}$

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

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