Question

Find the integrals of the function: $\frac{cos\ x}{1+ cos\ x}$

Answer 1

Answer:

Step-by-step explanation:

$Concept:\\\\cosA=2cos^2\frac{A}{2}-1\\\\Now,\\\\\int{\frac{cosx}{1+cosx}}\:dx\\\\=\int{\frac{(1+cosx)-1}{1+cosx}}\:dx\\\\=\int[\frac{1+cosx}{1+cosx}-\frac{1}{1+cosx}]\:dx\\\\=\int[1-\frac{1}{1+cosx}]\:dx\\\\=\int[1-\frac{1}{2cos^2\frac{x}{2}}]\:dx\\\\=\int[1-\frac{1}{2}sec^2\frac{x}{2}]\:dx\\\\=\int{1}\:dx-\frac{1}{2}\int{sec^2\frac{x}{2}}\:dx\\\\=x-\frac{1}{2}(\frac{tan\frac{x}{2}}{\frac{1}{2}})+c\\\\=x-tan\frac{x}{2}}+c$