Question

Find the integrals of the function: $\frac{sin2 x}{1+cos x}$

Answer 1

$\frac{sin2 x}{1+cos x}$ $= \frac{(2sin (\frac{x}{2}) cos \frac{x}{2})^2 }{2cos^2 (\frac{x}{2})}$ $[sin x = 2sin (\frac{x}{2}) cos (\frac{x}{2}) ; cos x = 2cos^2 (\frac{x}{2}) -1]$

$= \frac{4sin^2 \frac{x}{2} cos^2 (\frac{x}{2}) }{ 2cos^2\frac{x}{2}}$

$= 2sin^2 (\frac{x}{2})$
= 1-cos x

∴ ∫$\frac{sin2 x}{1+cos x}$ dx = ∫(1-cos x)dx

= x - sin x+C