Question

If \(\int_0^{\pi/4} \left[ \cot(x - \frac{\pi}{3}) - \cot(x + \frac{\pi}{3}) + 1 \right] dx = \alpha \log(\sqrt{3} - 1)\) then \(9\alpha\) is

• A. 12
• B. 14
• C. 9
• D. 17

Answer 1

The Correct Option is C

Step 1: Applying the Cotangent Subtraction Formula:
Recall the identity: \[ \cot(A-B) = \frac{\cot A\cot B + 1}{\cot B - \cot A} \] Therefore: \[ \cot(A-B)(\cot B - \cot A) = \cot A\cot B + 1 \] In our integral, let $A = x + \dfrac{\pi}{3}$ and $B = x - \dfrac{\pi}{3}$, so $A - B = \dfrac{2\pi}{3}$.
We have: \[ \cot\!\left(x-\frac{\pi}{3}\right)\cot\!\left(x+\frac{\pi}{3}\right) + 1 = \cot\!\left(\frac{2\pi}{3}\right)\left[\cot\!\left(x-\frac{\pi}{3}\right) - \cot\!\left(x+\frac{\pi}{3}\right)\right] \cdot(-1) \] More directly, from the identity $\cot(A-B) = \dfrac{\cot A \cot B + 1}{\cot B - \cot A}$: \[ \cot A\cot B + 1 = \cot(A-B)(\cot B - \cot A) \] Here with $A = x+\dfrac{\pi}{3}$, $B = x-\dfrac{\pi}{3}$: \[ \cot\!\left(x+\frac{\pi}{3}\right)\cot\!\left(x-\frac{\pi}{3}\right) + 1 = \cot\!\left(\frac{2\pi}{3}\right)\left[\cot\!\left(x-\frac{\pi}{3}\right) - \cot\!\left(x+\frac{\pi}{3}\right)\right] \] Since $\cot\!\left(\dfrac{2\pi}{3}\right) = -\dfrac{1}{\sqrt{3}}$, we get: \[ I = \int_{\pi/6}^{\pi/4} \frac{-1}{\sqrt{3}}\left[\cot\!\left(x-\frac{\pi}{3}\right) - \cot\!\left(x+\frac{\pi}{3}\right)\right]dx \] Step 2: Integrating:
Using $\displaystyle\int \cot u\,du = \ln|\sin u|$: \[ I = \frac{-1}{\sqrt{3}}\left[\ln\!\left|\sin\!\left(x-\frac{\pi}{3}\right)\right| - \ln\!\left|\sin\!\left(x+\frac{\pi}{3}\right)\right|\right]_{\pi/6}^{\pi/4} \] \[ = \frac{-1}{\sqrt{3}}\left[\ln\!\left|\frac{\sin\!\left(x-\frac{\pi}{3}\right)}{\sin\!\left(x+\frac{\pi}{3}\right)}\right|\right]_{\pi/6}^{\pi/4} \] Step 3: Evaluating at the Limits:
At $x = \dfrac{\pi}{4}$: $x - \dfrac{\pi}{3} = -\dfrac{\pi}{12}$, $x + \dfrac{\pi}{3} = \dfrac{7\pi}{12}$.
$\sin\!\left(-\dfrac{\pi}{12}\right) = -\sin 15°$, $\sin\!\left(\dfrac{7\pi}{12}\right) = \sin 75°$.
Ratio: $\left|\dfrac{-\sin 15°}{\sin 75°}\right| = \dfrac{\sin 15°}{\sin 75°} = \tan 15°$ (since $\sin 75° = \cos 15°$).
At $x = \dfrac{\pi}{6}$: $x - \dfrac{\pi}{3} = -\dfrac{\pi}{6}$, $x + \dfrac{\pi}{3} = \dfrac{\pi}{2}$.
$\sin\!\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2}$, $\sin\!\left(\dfrac{\pi}{2}\right) = 1$.
Ratio: $\dfrac{1/2}{1} = \dfrac{1}{2}$.
Step 4: Computing the Integral:
\[ I = \frac{-1}{\sqrt{3}}\left[\ln(\tan 15°) - \ln\!\left(\frac{1}{2}\right)\right] \] \[ = \frac{-1}{\sqrt{3}}\left[\ln(2-\sqrt{3}) + \ln 2\right] \] (since $\tan 15° = 2 - \sqrt{3}$)
\[ = \frac{-1}{\sqrt{3}}\ln\!\left(2(2-\sqrt{3})\right) = \frac{-1}{\sqrt{3}}\ln\!\left(4-2\sqrt{3}\right) \] Now $4 - 2\sqrt{3} = (\sqrt{3}-1)^2$, so: \[ I = \frac{-1}{\sqrt{3}}\cdot 2\ln(\sqrt{3}-1) = \frac{-2}{\sqrt{3}}\ln(\sqrt{3}-1) \] Given $I = \alpha\ln(\sqrt{3}-1)$: \[ \alpha = \frac{-2}{\sqrt{3}} \] \[ 9\alpha^2 = 9 \cdot \frac{4}{3} = 12 \] Step 5: Final Answer:
\[ 9\alpha^2 = 12 \] The answer is Option (1).