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Evaluate the following $\int\frac{\cos x-\cos 2x}{1-\cos x}dx$

$\begin{array}{1 1} (A)\;x-2 \sin x +c \\(B)\;-x+2 \sin x +c \\(C)\;x+2 \sin x +c \\ (D)\;-x-2 \sin x +c \end{array} $

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  • $\int \cos x dx=\sin x+c$
  • $\cos 2x= 2\cos ^2 x-1$
Let $I=\large\int\frac{\cos x-\cos 2x}{1-\cos x}$$dx$
we know $\cos 2x= 2 \cos ^2 x-1$
Therefore $I=\int\large\frac{\cos x-(-2\cos^ 2x-1)}{1-\cos x}$$dx$
$=\int\large\frac{\cos x-2\cos^ 2x+1}{1-\cos x}dx$
On factorizing $-2 \cos ^2 x+\cos x+1$, we get
$(2 \cos x-1)(\cos x-1)$
Therefore $I= \large\int \frac {(2 \cos x+1)(1-\cos x)}{(1-\cos x)}$$dx$
$=\int (2 \cos x+1)dx$
On integrating we get
$x+2 \sin x +c$
answered Apr 16, 2013 by meena.p
edited Apr 11 by meena.p
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