Question

$\int \frac{cosx-cos2x}{1-cosx}dx=$

Answer 1

$\int \frac{cosx-cos2x}{1-cosx}dx=\int \frac{2sin\frac{3x}{2}.sin\frac{x}{2}}{1-1+2si{n}^2\frac{x}{2}}dx=2\int \frac{sin\frac{3x}{2}.sin\frac{x}{2}}{2si{n}^2\frac{x}{2}}dx=\int \frac{sin\frac{3x}{2}}{sin\frac{x}{2}}dx=\int \frac{3sin\frac{x}{2}-4si{n}^3\frac{x}{2}}{sin\frac{x}{2}}dx[\therefore sin3x=3sinx-4si{n}^{3}x]=3\int dx-4\int si{n}^2\frac{x}{2}dx=3\int dx-4\int \frac{1-cosx}{2}dx=3\int dx-2\int dx+2\int cosxdx=\int dx+2\int cosxdx=x+2sinx+C=2sinx+x+C$