Ask Questions, Get Answers

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

$\Large \int\frac{\cos 2x-\cos 2\theta}{\cos x-\cos\theta}\normalsize dx$ is equal to\begin{array}{1 1}(A)\;2(\sin x+x\cos\theta)+C & (B)\;2(\sin x-x\cos\theta)+C\\(C)\;2(\sin x+2x\cos\theta)+C & (D)\;2(\sin x-2x\cos\theta)+C\end{array}

Can you answer this question?

1 Answer

0 votes
  • $\cos 2x=2\cos^2x-1$
  • $\int \cos x dx=\sin x+c$
Step 1:
$I=\Large \int\frac{\cos 2x-\cos 2\theta}{\cos x-\cos\theta}\normalsize dx$
$\cos 2x=2\cos^2x-1$
$I=\Large \int\frac{(2\cos ^2x-1)-(2\cos^2\theta-1)}{\cos x-\cos\theta}\normalsize dx$
On simplifying we get,
$\;\;\Large \int\frac{2(\cos ^2x-cos^2\theta)}{\cos x-\cos\theta}\normalsize dx$
$(\cos^2x-\cos^2\theta)=(\cos x+\cos \theta)(\cos x-\cos \theta)$
$\;\;=\int\large \frac{2(\cos x+\cos \theta)(\cos x-\cos\theta)}{\cos x-\cos \theta}\normalsize dx.$
$\;\;=2\int (\cos x+\cos\theta)dx.$
Step 2:
This can be splitted as
$I=2\begin{bmatrix}\int \cos x dx+\int \cos \theta.dx\end{bmatrix}$
On integrating we get,
$2(\sin x+x\cos\theta)+c$
Hence the correct option is (A)
answered Apr 23, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App