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What is the integration of \[ \int {\frac{{\cos x - \cos 2x}}{{1 - \cos x}}} dx\] ~~~~~[NCERT Exemp. Q. 25,Page 165 ] - Best E-books, Mathematics, Astrology, Sample Papers - CBSE, ISC, ICSE, JEE, BITSAT & SAT
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What is the integration of \[ \int {\frac{{\cos x – \cos 2x}}{{1 – \cos x}}} dx\] ~~~~~[NCERT Exemp. Q. 25,Page 165 ]

Let \[ I = \int {\frac{{\cos x - \cos 2x}}{{1 - \cos x}}} dx = \int {\frac{{2\sin \frac{{3x}}{2} \cdot \sin \frac{x}{2}}}{{1 - 1 + 2{{\sin }^2}\frac{x}{2}}}} dx\]


\[ = 2\int {\frac{{\sin \frac{{3x}}{2} \cdot \sin \frac{x}{2}}}{{2{{\sin }^2}\frac{x}{2}}}} dx = \int {\frac{{\sin \frac{{3x}}{2}}}{{\sin \frac{x}{2}}}} dx\]

\[ = \int {\frac{{3\sin \frac{x}{2} - 4{{\sin }^3}\frac{x}{2}}}{{\sin \frac{x}{2}}}} dx\]


\[ = 3\int d x - 4\int {{{\sin }^2}} \frac{x}{2}dx = 3\int d x - 4\int {\frac{{1 - \cos x}}{2}} dx\]

\[ = 3\int d x - 2\int d x + 2\int {\cos } xdx\]


\[ = \int d x + 2\int {\cos } xdx = x + 2\sin x + C = 2\sin x + x + C\]

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