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\[\frac{\cos x-\sin x}{1+\sin2x}\]

Can you answer this question?

1 Answer

0 votes
  • $(i)\sin^2x+\cos^2x=1.$
  • $(ii)\sin2x=2\sin x\cos x.$
  • $(iii)method\;of\;substitution:$
  • If f(x)=t.Then f'(x)dx=dt.
  • Thus $I=\int f(x)dx=\int t.dt.$
Given $I=\int\frac{\cos x-\sin x}{1+\sin2x}dx.$
This can be written as,(since $\sin^2x+\cos^2x=1)$
$I=\int\frac{\cos x-\sin x}{\sin^2x+\cos^2x+2\sin x\cos x}dx.$
$\;\;\;=\int\frac{\cos x-\sin x}{(\sin x+\cos x)^2}dx.$
Put $\sin x+\cos x=t.$
On differentiating we get,
$(\cos x-\sin x)dx=dt.$
Now substituting t and dt we get,
On integrating we get,
Substituting for t we get,
$\;\;\;=-\frac{1}{(\sin x+\cos x)}+c.$
$\;\;\;=\frac{-1}{(\sin x+\cos x)}+c.$


answered Jan 31, 2013 by sreemathi.v
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