How does (1/2)∫[-π/2,π/2(16+9sin^2(θ))dθ become (1/2)*2∫[0,π/2(16+9(1/2)(1-cos(2θ)))dθ? Where does the 1/2 between +9 and 1-cos(2θ come from and how does sin^2(θ) become cos(2θ) instead of cos^2(θ)?

Answer 1

the tricky part is to integrate sin^2 x

recall that cos (2x) = 1 - 2sin^2 x
2 sin^2 x = 1 - cos (2x)
sin^2 x = 1/2(1 - cos (2x) )

so in the (1/2)∫[-π/2,π/2(16+9sin^2(θ))dθ
you will see :
(1/2)∫[-π/2,π/2(16+9(1/2)(1 - cos (2x) ))dθ

You should be able to integrate the cos (2x) part without any difficulty

Answer 2

small typo:
I changed θ to x in my calculation, then forgot to change it back,
but I am sure that did not bother you.