JEE Main 2024 (Online) 1st February Evening ShiftQuestion 1
If $\int_{0}^{\pi/3}\cos^{4}x\,dx=a\pi+b\sqrt{3}$, where $a$ and $b$ are rational numbers, then $9a+8b$ is equal to :
Definite Integration
Solution
Concept. Write $\cos^4x$ as a sum of cosines of multiple angles and integrate term by term. Formulas used. (1) $\cos^4x=\Big(\frac{1+\cos2x}2\Big)^2=\frac38+\frac12\cos2x+\frac18\cos4x$Step 1 - Integrate. By (1), $$\int_0^{\pi/3}\cos^4x\,dx=\frac38\cdot\frac\pi3+\frac14\sin\frac{2\pi}3+\frac1{32}\sin\frac{4\pi}3.$$ Step 2 - Values. $\sin\frac{2\pi}3=\frac{\sqrt3}2$, $\sin\frac{4\pi}3=-\frac{\sqrt3}2$: $$\frac\pi8+\frac{\sqrt3}8-\frac{\sqrt3}{64}=\frac\pi8+\frac{7\sqrt3}{64}.$$ So $a=\frac18$, $b=\frac7{64}$. Step 3 - Value. $9a+8b=\frac98+\frac78=2$. Answer: (B)