By first proving that sin2θ=2sinθcosθ, calculate ∫1+sinθcosθ dθ.

We have, from the formula book, sin⁡(A±B)=sinAcosB±cosAsinB Using A=B=θ, we have sinθ+θ=sinθcosθ+cosθsinθ Which we can simplify to sin2θ=2sinθcosθ as required. We can then substitute this into the integral: 1+1/2sin2θ dθ From this we can calculate the integral, 1+1/2sin2θ dθ =θ-1/4cos2θ+c where c is an arbitrary constant.