# Find the integrals of the functions$\frac{\cos2x}{(\cos x+\sin x)^2}$

$\begin{array}{1 1}\log | \sin x+\cos x |+c. \\ \log | \sin x-\cos x |+c \\ \log | 2\sin x+2\cos x |+c \\ \log | 2\sin x-\cos x |+c. \end{array}$

Toolbox:
• $(i)\;\sin2x=2\sin x\cos x.$
• $(ii)\;\int\frac{1}{x}dx=log x+c.$
• $(iii)\;Method\;of\;substitution:$
• If f(x)=t.Therefore f'(x)dx=dt.Thus $\int f(x)dx=\int t.dt.$
Given:$I=\int\frac{\cos 2x}{(\cos x+\sin x)^2}dx.$

$\;\;\;=\int\frac{\cos 2x}{\cos^2+\sin^2x+2\sin x\cos x}dx.$

But $2\sin x\cos x=\sin 2x$ and $\sin^2x+\cos^2x=1.$

$\;\;\;=\int\frac{\cos 2x}{1+\sin2x}dx.$

Let $1+\sin 2x=t.$

On differentiating

$2\cos 2xdx=dt.$

On substituting t we get,

$I=\int\frac{dt/2}{t}=\frac{1}{2}\int\frac{1}{t}.dt.$

$\sec^2xdx=dt.$

By substituting t we get,

$I=\int t.dt+\int\frac{1}{t}.dt.$

On integrating we get,

$\;\;\;=\frac{1}{2}log |t|+c.$

On Substituting back for t we get,

$\;\;\;=\frac{1}{2}log|1+\sin 2x|+c.$

But $\sin^2x+\cos^2x=1$ and $\sin2x=2\sin x\cos x$.

$\;\;\;=\frac{1}{2}log|\sin^2x+\cos^2x+2\sin x\cos x|+c.$

This is of the form $(a+b)^2$

$\;\;\;=\frac{1}{2}log(\sin x+\cos x)^2+c.$

$\frac{1}{2}log(\sin x+\cos x)^2$can be written as $log[(\sin x\tan x)^2]$

$\;\;\;=log\sqrt {(\sin x+\cos x)^2}+c.$

$\;\;\;=log|\sin x+\cos x|+c.$

edited Jul 21, 2013