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Find the integrals of the functions\[\frac{\cos x-\sin x}{1+\sin2x}\]

1 Answer

  • $(i)\sin^2x+\cos^2x=1.$
  • $(ii)\sin2x=2\sin x\cos x.$
  • $(iii)method\;of\;substitution:$
  • 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