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# Evaluate:$\int \limits_0^{\pi /2} \frac{1}{\cos (x-\pi/3) \cos (x-\pi/6)}$

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Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
• $\sin (A - B)=\sin A \cos B-\cos A \sin B$
• $\int \tan x dx =-log |\cos x|+c$
Given $\int \limits_0^{\pi /2} \frac{1}{\cos (x-\pi/3) \cos (x-\pi/6)}$

Multiply and divide by $\sin (\frac{\pi}{3}-\frac{\pi}{6})=\sin (\frac{\pi}{6})=\frac{1}{2}$

$\large\frac{1}{\sin(\frac{\pi}{3}-\frac{\pi}{6})} \int \limits_0^{\pi /2} \frac{\sin(\pi/3-\pi/6)}{\cos (x-\pi/3) \cos (x-\pi/6)}dx$

$=\large\frac{1}{1/2}\int \limits_0^{\pi /2} \frac{\sin[(x-\pi/6)-(x-\pi/3)]}{\cos (x-\pi/3) \cos (x-\pi/6)}dx$

we know $\sin (A-B)= \sin A \cos B-\cos A\sin B$

similarly,

$I=2\int \limits_0^{\pi /2} \large\frac{\sin(x-\pi/6) \cos (x-\pi/3)-\cos(x-\pi/6)\sin(x-\pi/3)}{\cos (x-\pi/3) \cos (x-\pi/6)}dx$

On seperating the terms

$=2\int \limits_0^{\pi /2}\large \frac{\sin(x-\pi/6) \cos (x-\pi/3)}{\cos (x-\pi/3) \cos (x-\pi/6)}-2\int \limits_0^{\pi /2} \large\frac{\cos(x-\pi/6)\sin(x-\pi/3)}{\cos (x-\pi/6) \cos (x-\pi/3)}dx$
$=2\int \limits_0^{\pi /2} \large\frac{\sin(x-\pi/6)}{\cos (x-\pi/6)}dx-2\int \limits_0^{\pi /2} \frac{\sin(x-\pi/3)}{\cos (x-\pi/3)}dx$

$=2\int \limits_0^{\pi /2} \tan (x-\pi/6)dx-2\int \limits_0^{\pi /2} \tan (x-\pi/3)dx$

But we know that $\int \tan x=-log \cos |x|$

on integrating

Therefore $I=2[-log \cos |x-\pi/6|]_0^{\pi/2}-2[-log(x-\pi/3)]_0^{\pi/2}$

$=2\bigg\{[-log \cos |x-\pi/3|]_0^{\pi/2}-[log \cos|x-\pi/6|_0^{\pi/2}\bigg\}$

$=2 \bigg[log \bigg|\large\frac{\cos(x-\pi/3)}{\cos (x-\pi/6)}\bigg|\bigg]_0^{\pi /2}$

on applying limits,

$2 \bigg\{\bigg[log \bigg(\large\frac{\cos(\pi/2-\pi/3)}{\cos (\pi/2-\pi/6)}\bigg)\bigg]-log \bigg( \frac{\cos(0-\pi/3)}{\cos (0-\pi/6)}\bigg)\bigg\}$

$=2 \bigg[log \large\frac{\cos(\pi/6)}{\cos (\pi/3)}\bigg]-log \bigg[\frac{\cos \pi/3}{\cos \pi/6}\bigg]$

But we know $\cos (\pi/6)=\frac{\sqrt 3}{2}$ and $\cos (\pi/3)=\frac{1}{2}$

Therefore $I=2 \bigg[ log \frac{\sqrt 3/2}{1/2}\bigg]-\bigg[log \frac{1/2}{\sqrt 3/2}\bigg]$

$=2 \bigg\{log \frac{\sqrt 3/2}{1/2} \times \frac {\sqrt 3/2}{1/2}\bigg\}$

$=2 log 3$