Question

Find the integrals of the function: $\frac{1-cosx}{1+cosx}$

Answer 1

$\frac{1-cosx}{1+cosx}$ =$\frac{ 2sin^2 \frac x2}{2cos^2 \frac x2}$t $[2sin^2\frac{x}{2}=1-cosx\, and \,2cos^2\frac{x}{2}=1+cosx$
=tan2$\frac x2$
=(sec2 $\frac x2$-1)
∴ ∫$\frac{1-cosx}{1+cosx}dx$= ∫(sec2 $\frac x2$-1)dx
=[$\frac {tan(\frac {x}{2})}{\frac 12}$-x]+C
=2 tan$\frac x2$-x+C