logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

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

Can you answer this question?
 
 

1 Answer

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

 

 

answered Mar 15, 2013 by meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...