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

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

$I=\int\frac{dt}{t^2}.$

On integrating we get,

$\;\;\;=\frac{t^{-1}}{-1}+c.$

$\;\;\;=\frac{-1}{t}+c.$

Substituting for t we get,

$\;\;\;=-\frac{1}{(\sin x+\cos x)}+c.$

$\;\;\;=\frac{-1}{(\sin x+\cos x)}+c.$

answered Jan 31, 2013