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)?

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

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.

To understand how the given integral expression transforms from (1/2)∫[-π/2,π/2](16+9sin^2(θ))dθ to (1/2)*2∫[0,π/2](16+9(1/2)(1-cos(2θ)))dθ, let's break it down step by step:

Step 1: Expand sin^2(θ)
In trigonometry, we have the identity sin^2(θ) = (1/2)(1 - cos(2θ)). This formula relates the square of the sine function to the cosine function. This expansion is derived using the double-angle formula for cosine, which states that cos(2θ) = 1 - 2sin^2(θ).

Step 2: Applying the expansion
Replace sin^2(θ) in the original integral expression with its expanded form: (1/2)(1 - cos(2θ)).
(1/2)∫[-π/2,π/2](16 + 9(1/2)(1 - cos(2θ)))dθ.

Step 3: Simplify the integral expression
Distribute the factor of 9/2 to both terms inside the parentheses:
(1/2)∫[-π/2,π/2](16 + (9/2)(1 - cos(2θ)))dθ.

Step 4: Simplify further
Multiply (9/2)(1 - cos(2θ)) to get (9/2) - (9/2)cos(2θ):
(1/2)∫[-π/2,π/2](16 + (9/2) - (9/2)cos(2θ))dθ.

Step 5: Combine like terms
Combine the constant terms:
(1/2)∫[-π/2,π/2](25/2 - (9/2)cos(2θ))dθ.

Step 6: Adjust the limits of integration
Since the function being integrated is even (symmetric about the y-axis), we can change the limits of integration from [-π/2, π/2] to [0, π/2]:
(1/2)∫[0,π/2](25/2 - (9/2)cos(2θ))dθ.

Further simplification to the expression can be done, but the primary step was to explain the substitution made from sin^2(θ) to cos(2θ) and the presence of the factor 1/2.