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# Choose the correct answers in $\int\frac{\large \cos 2x}{\large (\sin x+\cos x)^2}dx$ is equal to

$\begin{array}{1 1} A\;\frac{-1}{\sin x+\cos x}+C \\ (B)\;log|\sin x+\cos x\;|+C \\ C\;log|\sin x-\cos x\;|+C \\D\;\frac{1}{(\sin x+\cos x)^2} \end{array}$

Toolbox:
• (i)In a function f(x) is substituted by t, then $f'(x)dx=dt$ then $\int f(x)dx=\int t.dt$
• (ii) $\cos 2x=\cos ^2 x-\sin ^2 x$
• (iii) $\int \frac{dx}{x}=log |x|+c$
Given $I=\int \frac{\cos 2x}{(\cos x+\sin x )^2}dx$

But we know $\cos 2x=\cos ^2x-\sin ^2x$

Therefore $I=\int \frac{\cos ^2x -\sin ^2x}{(\cos x+\sin x)^2}$

$\cos ^2x-\sin ^2 x=(\cos x+\sin x)(\cos x-\sin x)$

Therefore $I=\int \frac{(\cos x+\sin x)(\cos x -\sin x)}{(\cos x+\sin x)^2}$

$=\int \frac{\cos x-\sin x}{\cos x+\sin x} dx$

Let $\cos x +\sin x=t;$ on differentiating w.r.t x we get

$(-\sin x+\cos x)dx=dt$

Substituting for t and dt

$I=\int \frac{dt}{t}$

on integrating we get

$I=log |t|+c$

substituting for t we get

$I=log |\cos x +\sin x|+c$

Hence the correct answer is B