Question

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

Answer 1

$\frac{cos x}{1+cos x}$ = $\frac{cos^2 (\frac{x}{2}) -sin^2 (\frac{x}{2}) }{2cos^2 (\frac{x}{2}) }$ [cos x= cos2 $\frac{x}{2}$-sin2 $\frac{x}{2}$ and cos x = 2cos2 $\frac{x}{2}$-1]
=$\frac{1}{2}$[1-tan2 $\frac{x}{2}$]
∴ ∫$\frac{cos x}{1+cos x}dx$ = $\frac{1}{2}$∫(1-tan2$\frac{x}{2}$)dx
= $\frac{1}{2}$ ∫(1-sec2 $\frac{x}{2}$+1)dx
= $\frac{1}{2}$∫(2-sec2 $\frac{x}{2}$)dx
= $\frac{1}{2}$[$2x-\frac{tan(\frac{x}{2})}{(\frac{1}{2})}$]+C
=x -tan$\frac{x}{2}$+C