If ∫0π3cos4x dx=aπ+b3, where a and b are rational numbers, Then 9a+8b is equal to: [2024]
(1)
Let I=∫0π/3cos4x dx=∫0π/3(1+cos2x2)2 dx
=14∫0π/3(1+cos22x+2cos2x) dx
=14∫0π/3(1+2cos2x+(1+cos4x2)) dx
=14∫0π/3(32+2cos2x+cos4x2) dx
=14[32x+sin2x+sin4x8]0π/3
=14[π2+32-316]=π8+7364⇒a=18, b=764
∴ 9a+8b=98+78=2