Question

Find the integrals of the function: $\frac{cos x - sin x}{1+sin 2x}$

Answer 1

$\frac{cos x - sin x}{1+sin 2x}$ $= \frac{cos x -sin x}{(sin^2 x + cos^2 x)+2sin x cos x }$ [sin2 x+cos2 x =1; sin 2x=2sin x cos x]

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

Let $sin x +cos x = t$

$∴ (cos x-sin x)dx = dt$

⇒ ∫$\frac{cos x - sin x}{1+sin 2x}$ = ∫$\frac {cos x-sin x}{(sin x + cos x)^2}$ dx

$= ∫\frac{dt}{t^2}$

$= ∫t^2dt$

$=-t^2+C$

$=-\frac{1}{t}+C$

$=\frac{-1}{sin x +cos x} +C$