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.
AH
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