Q.388. Definite integration.

Question 388.
Evaluate ∫ (x = 0 to π) (a^2 cos^2 x + b^2 sin^2 x) dx

Answer 388.
Using the formula
∫ (x = 0 to π) f (x) dx = 2 ∫ (x = 0 to π/2) f (x) dx, if f (π - x) = f (x),
=> ∫ (x = 0 to π) (a^2 cos^2 x + b^2 sin^2 x) dx
= 2 ∫ (x = 0 to π/2) (a^2 cos^2 x + b^2 sin^2 x) dx ... ( 1 )
= 2 ∫ (x = 0 to π/2) [a^2 cos^2 (π/2 - x) + b^2 sin^2 (π/2 - x)] dx
.........................Using ∫ (x = 0 to a) f (x) dx = ∫ (x = 0 to a) f (a - x) dx]
= 2 ∫ (x = 0 to π/2) (a^2 sin^2 x + b^2 cos^2 x) dx ... ( 2 )

Adding ( 1 ) and ( 2 ),
2 ∫ (x = 0 to π) (a^2 cos^2 x + b^2 sin^2 x) dx
= 2 ∫ (x = 0 to π/2) [a^2 (cos^2 x + sin^2 x) + b^2 (sin^2 x + cos^2 x)] dx
=> ∫ (x = 0 to π) (a^2 cos^2 x + b^2 sin^2 x) dx
= ∫ (x = 0 to π/2) (a^2 + b^2) dx
= (a^2 + b^2) [x] ... (x = 0 to π/2)
= (π/2) (a^2 + b^2).

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