The integral _/6^/4(dx)/(2x(^5x+^5x)) equals:

Question

The integral $\displaystyle\int_{\pi/6}^{\pi/4}\frac{dx}{\sin2x(	an^5x+\cot^5x)}$ equals:

Accepted Answer

$$I=\int_{\pi/6}^{\pi/4}\frac{dx}{2\sin x\cos x\left(\tan^5x+\dfrac1{\tan^5x}\right)}=\frac12\int_{\pi/6}^{\pi/4}\frac{\tan^4x\sec^2x}{(\tan^5x)^2+1}dx$$

Let $\tan^5x=t\Rightarrow5\tan^4x\sec^2x\,dx=dt$; limits: $x=\dfrac\pi6\to t=\dfrac1{3\sqrt3\cdot9}=\dfrac1{9\sqrt3\cdot\sqrt3}\cdots$, more precisely $t=\left(\dfrac1{\sqrt3}\right)^5=\dfrac1{9\sqrt3}$, and $x=\dfrac\pi4\to t=1$:

$$I=\frac1{10}\int_{1/(9\sqrt3)}^1\frac{dt}{t^2+1}=\frac1{10}\left[\tan^{-1}t\right]_{1/(9\sqrt3)}^1=\frac1{10}\left(\frac\pi4-\tan^{-1}\frac1{9\sqrt3}\right)$$

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