Ask Questions, Get Answers

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

Find the integrals of the functions\[\frac{1}{\cos(x-a)\cos(x-b)}\]

$\begin{array}{1 1} \frac{log\mid \frac{\cos(x-a)}{\cos(x-b)}\mid}{\sin(a-b)}+c \\ \frac{log\mid \frac{\cos(x-a)}{\sin (x-b)}\mid}{\sin(a-b)}+c \\ \frac{log\mid \frac{\cos(x-a)}{\cos(x-b)}\mid}{\cos (a-b)}+c \\ \frac{log\mid \frac{\sin (x-a)}{\cos(x-b)}\mid}{\cos (a-b)}+c\end{array}$

Can you answer this question?

1 Answer

0 votes
  • $(i)\;\sin(A+B)=\sin A\cos B+\cos A\sin B.$
  • $(ii)\;\int\tan xdx=-log|\cos x|+c.$
Given:$I=\int\frac{1}{\cos (x-a)\cos(x-b)}dx.$
Multiply and divide by $\sin(a-b)$
$I=\frac{1}{\sin(a-b)}\int\frac{\sin(a-b)}{\cos (x-a)\cos(x-b)}dx.$
Add and subtract x in $\sin(a-b)$
On expanding we get,
Now separating the terms we get,
On integrating we get,
$log a-log b=log\mid\frac{a}{b}\mid.$
In the same way we get,


answered Feb 3, 2013 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App