Question

\( \int \frac{\cos x-\cos 2 x}{1-\cos x} d x \)

Answer 1

∵ \(\frac{cos\,x-cos\,2x}{1-cos\,x}=\frac{cos\,x-(2cos^2x-1)}{1-cos\,x}\) (∵ cos 2x = 2 cos²x - 1)

\(=\frac{-2cos^2x+cos\,x+1)}{1-cos\,x}\)

\(=\frac{-2cos^2x+2\,cos\,x-cos\,x+1)}{1-cos\,x}\)

\(=\frac{2\,cos\,x(1-cos\,x)+(1-cos\,x)}{1-cos\,x}\)

\(=\frac{(1-cos\,x)(2\,cos\,x+1)}{1-cos\,x}\)

= 2 cos x+1

∴ ∫\(\frac{cos\,x-cos\,2x}{1-cos\,x}dx\) = ∫(2 cos x+1)dx

= 2∫cos x dx + ∫dx

= 2 sin x+x+c