If _ /6^ /4( (x-( )/(3)) (x+( )/(3))+1)dx= _e(√3-1), then 9 ^2 is equal to ____ .

Question

If $\int _{\pi /6}^{\pi /4}(\cot (x-\frac{\pi }{3})\cot (x+\frac{\pi }{3})+1)dx=\alpha {\log }_{e}(\sqrt{3}-1)$, then $9{\alpha }^{2}$ is equal to $____$ .

Answer: 12

Accepted Answer

$I=\int _{\frac{\pi }{6}}^{\frac{\pi }{4}}[\cot (x-\frac{\pi }{3})\cdot \cot (x+\frac{\pi }{3})+1]dx$

$\begin{gathered} ={\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}|\frac{\cos (x-\frac{\pi }{3})\cdot \cos (x+\frac{\pi }{3})+\sin (x-\frac{\pi }{3})\cdot \sin (x+\frac{\pi }{3})}{\sin (x-\frac{\pi }{3})\cdot \sin (x+\frac{\pi }{3})}|dx \ ={\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}|\frac{\cos {(x+\frac{\pi }{3})-(x-\frac{\pi }{3})}}{\sin (x-\frac{\pi }{3})\cdot \sin (x+\frac{\pi }{3})}|dx \ =\cos (\frac{2\pi }{3})\cdot {\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}[\frac{1}{\sin (x-\frac{\pi }{3})\cdot \sin (x+\frac{\pi }{3})}]dx \end{gathered}$

$\Rightarrow$ $I=2\cos (\frac{2\pi }{3})\cdot {\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}[\frac{1}{2\sin (x-\frac{\pi }{3})\cdot \sin (x+\frac{\pi }{3})}]dx$

$\begin{gathered} =2\cos (\frac{2\pi }{3})\cdot {\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}[\frac{1}{\cos (x-\frac{\pi }{3})\cdot \cos (x+\frac{\pi }{3})+\sin (x-\frac{\pi }{3})\cdot \sin (x+\frac{\pi }{3})}-{\cos (x-\frac{\pi }{3})\cdot \cos (x+\frac{\pi }{3})-\sin (x-\frac{\pi }{3})\cdot \sin (x+\frac{\pi }{3})]}dx \ =2\cos (\frac{2\pi }{3})\cdot {\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}[\frac{1}{\cos (\frac{2\pi }{3})-\cos (2x)}]dx \end{gathered}$

$\Rightarrow$ $I=-{\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}[\frac{1}{-\frac{1}{2}-\cos (2x)}]dx={\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}[\frac{2}{1+2\cos (2x)}]dx={\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}[\frac{2}{1+2(1-2{\sin }^{2}x)}]dx={\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}[\frac{2}{3-4{\sin }^{2}x)}]dx$

$\Rightarrow$ $I={\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}[\frac{2{\sec }^{2}x}{3{\sec }^{2}x-4{	an }^{2}x}]dx={\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}[\frac{2{\sec }^{2}x}{3(1+{	an }^{2}x)-4{	an }^{2}x}]dx={\int }_{\frac{\pi }{4}}^{\frac{\pi }{6}}[\frac{2{\sec }^{2}x}{3-{	an }^{2}x}]dx$

Substitute $t=	an x\Rightarrow dt=({\sec }^{2}x)dx$

$\Rightarrow$ $I={\int }_{1}^{\frac{1}{\sqrt{3}}}[\frac{2}{3-{t}^{2}}]dt=(\frac{1}{\sqrt{3}})\cdot {\int }_{1}^{\frac{1}{\sqrt{3}}}[\frac{2\sqrt{3}}{(\sqrt{3}+t)(\sqrt{3}-t)}]dt$

$=(\frac{1}{\sqrt{3}})\cdot {\int }_{1}^{\frac{1}{\sqrt{3}}}[\frac{(\sqrt{3}+t)+(\sqrt{3}-t)}{(\sqrt{3}+t)(\sqrt{3}-t)}]dt=(\frac{1}{\sqrt{3}})\cdot {\int }_{1}^{\frac{1}{\sqrt{3}}}[\frac{1}{\sqrt{3}-t}+\frac{1}{\sqrt{3}+t}]dt$

$\Rightarrow$ $I=(\frac{1}{\sqrt{3}})\cdot [\ln (\sqrt{3}+t)-\ln (\sqrt{3}-t){]}_{1}^{\frac{1}{\sqrt{3}}}=(\frac{1}{\sqrt{3}})\cdot {[\ln (\frac{\sqrt{3}+t}{\sqrt{3}-t})]}_{1}^{\frac{1}{\sqrt{3}}}$

$=(\frac{1}{\sqrt{3}})\cdot [\ln (\frac{\sqrt{3}+1}{\sqrt{3}-1})-\ln 2]=(\frac{1}{\sqrt{3}})\cdot [\ln {\frac{2}{(\sqrt{3}-1{)}^{2}}}-\ln 2]$

$I=(\frac{1}{\sqrt{3}})\cdot \ln {\frac{2}{2(\sqrt{3}-1{)}^{2}}}=(\frac{1}{\sqrt{3}})\cdot \ln {\frac{1}{(\sqrt{3}-1{)}^{2}}}$

$=(\frac{1}{\sqrt{3}})\cdot \ln {(\sqrt{3}-1{)}^{-2}}=(-\frac{2}{\sqrt{3}})\cdot \ln (\sqrt{3}-1)\ldots \ldots (i)$

$	herefore $ $\alpha =-\frac{2}{\sqrt{3}}\Rightarrow 9{\alpha }^{2}=12$

Solutions · 1