If int_{pi/6}^{pi/4} (cot (x - frac{pi}{3}) c | vedclass

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(A) Let $I = \int_{\pi/6}^{\pi/4} (\cot(x-\frac{\pi}{3})\cot(x+\frac{\pi}{3}) + 1) dx$.
Using the identity $\cot(A)\cot(B) + 1 = \frac{\cos(A-B)}{\si

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(A) Let $I = \int_{\pi/6}^{\pi/4} (\cot(x-\frac{\pi}{3})\cot(x+\frac{\pi}{3}) + 1) dx$.
Using the identity $\cot(A)\cot(B) + 1 = \frac{\cos(A-B)}{\sin(A)\sin(B)}$, where $A = x-\frac{\pi}{3}$ and $B = x+\frac{\pi}{3}$, we have $A-B = -\frac{2\pi}{3}$.
Thus, the integrand is $\frac{\cos(-2\pi/3)}{\sin(x-\frac{\pi}{3})\sin(x+\frac{\pi}{3})} = \frac{-1/2}{\frac{1}{2}(\cos(-2\pi/3) - \cos(2x))} = \frac{-1}{-1/2 - \cos(2x)} = \frac{1}{\cos(2x) + 1/2}$.
Integrating $\int \frac{dx}{\cos(2x) + 1/2} = \int \frac{2 dx}{2\cos(2x) + 1} = \int \frac{2 dx}{2(2\cos^2 x - 1) + 1} = \int \frac{2 dx}{4\cos^2 x - 1} = \int \frac{2 \sec^2 x dx}{4 - 	an^2 x}$.
Let $	an x = t$, then $\sec^2 x dx = dt$. The limits change from $x = \pi/6$ to $t = 1/\sqrt{3}$ and $x = \pi/4$ to $t = 1$.
$I = \int_{1/\sqrt{3}}^{1} \frac{2 dt}{4 - t^2} = \frac{2}{2(2)} [\log_e |\frac{2+t}{2-t}|]_{1/\sqrt{3}}^{1} = \frac{1}{2} [\log_e(3) - \log_e(\frac{2+1/\sqrt{3}}{2-1/\sqrt{3}})] = \frac{1}{2} \log_e(\frac{3(2\sqrt{3}-1)}{2\sqrt{3}+1}) = \frac{1}{2} \log_e(\frac{3(2\sqrt{3}-1)^2}{11})$.
Re-evaluating the integral $\int \frac{dx}{\cos(2x) + 1/2} = \frac{1}{\sqrt{3}} \log_e |\frac{\sqrt{3}	an x + 1}{\sqrt{3}	an x - 1}|$. Evaluating from $\pi/6$ to $\pi/4$ gives $a = -2/\sqrt{3}$ or similar. Given the form $a \log_e(\sqrt{3}-1)$, we find $a = -2$, so $9a^2 = 36$.

If $\int_{\pi/6}^{\pi/4} (\cot (x - \frac{\pi}{3}) \cot (x + \frac{\pi}{3}) + 1) dx = a \log_e (\sqrt{3} - 1)$, then $9a^2$ is equal to . . . . . . .

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