Ask Questions, Get Answers

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

Integrate the function\[\int\frac{1}{\cos(x+a)\cos(x+b)}\]

Can you answer this question?

1 Answer

0 votes
  • (i)$\sin (A+B)=\sin A\cos B+\cos A\sin B$
  • (ii)$\cos (A+B)=\cos A\cos B-\sin A\sin B$
  • (iii)$\tan x dx=-log |\cos x|+c$
Given $\int\frac{dx}{\cos(x+a)\cos(x+b)}$
Multiply and divide by, $\sin (a-b)$
$I=\int\frac{\sin (a-b)\;dx}{\cos(x+a)\cos(x+b)(\sin (a-b))}$
Let $a-b=(x+a)-(x+b)$
$I=\large\frac{1}{\sin (a-b)}\int\frac{\sin[(x+a)-(x+b)]}{\cos(x+a)\cos(x+b)}dx$
But we know $\sin (A-B)=\sin A\cos B-\cos A\sin B$.
$\sin[(x+a)-(x+b)]$ is of the form $\sin(A-B),$ hence
$I=\large\frac{1}{\sin (a-b)}\int\frac{\sin(x+a)-\cos (x+b)-\cos (x+a)\sin (x+b)}{\cos(x+a)\cos(x+b)}dx$
Seperating the terms:
$=\large\frac{1}{\sin (a-b)}\int\frac{\sin(x+a)-\cos(x+b)}{\cos(x+a)\cos(x+b)}-\frac{\cos (x+a)\sin (x+b)}{\cos(x+a)\cos(x+b)}dx$
On simplifying we get,
$\large\frac{1}{\sin (a-b)}\bigg[\int\frac{\sin(x+a)}{\cos(x+a)}-{\sin (x+b)}{\cos(x+b)}dx\bigg]$
But $\large\frac{\sin(x+a)}{\cos(x+a)}\;and \;\frac{\sin (x+b)}{\cos(x+b)}=\tan{x+a}$
and $\tan (x+b)$ respectively
But $\frac{1}{\sin(a-b)}[-log|\cos(x+a)|+log|\cos(x+b)|+c]$
But we know $log a-log b=log \frac{a}{b}$
Therefore$ \int \frac{1}{\cos(x+a)\cos(x+b)}=\frac{1}{\sin(a-b)} log\bigg|\frac{\cos (x+b)}{\cos (x+a)}\bigg|+c$



answered Feb 17, 2013 by meena.p
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