JEE Main 20241 Feb, Evening shiftMathematicsHardIntegral Calculus369× since 2002If ∫0π3cos4x dx=aπ+b3\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x=\mathrm{a} \pi+\mathrm{b} \sqrt{3}0∫3πcos4x dx=aπ+b3, where a\mathrm{a}a and b\mathrm{b}b are rational numbers, then 9a+8b9 \mathrm{a}+8 \mathrm{b}9a+8b is equal to :AOption A. 2BOption B. 1COption C. 3DOption D. 32\frac{3}{2}23CheckSkip · show solution
