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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} $

1 Answer

  • (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



answered Mar 13, 2013 by meena.p