If \int\limits_0^\frac\pi3 \cos ^4 x ~d x=a \pi+b \sqrt3 , where a and b are rational numbers, then 9 a+8 b is equal to :

Definite Integration PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q36 with Solution

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Correct answer: A

\left(\frac{1+\cos 2x}{2}\right)^2.$$ So, $$\cos^4 x=\frac{1}{4}\left(1+2\cos 2x+\cos^2 2x\right).$$ Now, $$\cos^2 2x=\frac{1+\cos 4x}{2}.$$ Hence, $$\cos^4 x=\frac{1}{4}\left(1+2\cos 2x+\frac{1+\cos 4x}{2}\right) =\frac{1}{4}\left(\frac{3}{2}+2\cos 2x+\frac{1}{2}\cos 4x\right).$$ Thus, $$\cos^4 x=\frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x.$$ 3. Substitute into the integral: $$I=\int_0^{\pi/3}\left(\frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x\right)dx.$$ Integrate term by term: $$I=\left[\frac{3x}{8}+\frac{1}{2}\cdot\frac{\sin 2x}{2}+\frac{1}{8}\cdot\frac{\sin 4x}{4}\right]_0^{\pi/3}$$ $$=\left[\frac{3x}{8}+\frac{\sin 2x}{4}+\frac{\sin 4x}{32}\right]_0^{\pi/3}.$$ 4. Evaluate at the limits: At $x=\pi/3$, $$\frac{3x}{8}=\frac{3(\pi/3)}{8}=\frac{\pi}{8},$$ $$\sin\left(\frac{2\pi}{3}\right)=\frac{\sqrt{3}}{2},$$ $$\sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}.$$ So, $$I=\frac{\pi}{8}+\frac{1}{4}\cdot\frac{\sqrt{3}}{2}+\frac{1}{32}\cdot\left(-\frac{\sqrt{3}}{2}\right).$$ $$I=\frac{\pi}{8}+\frac{\sqrt{3}}{8}-\frac{\sqrt{3}}{64}.$$ $$I=\frac{\pi}{8}+\frac{8\sqrt{3}-\sqrt{3}}{64}$$ $$I=\frac{\pi}{8}+\frac{7\sqrt{3}}{64}.$$ 5. Compare with $$I=a\pi+b\sqrt{3}.$$ Therefore, $$a=\frac{1}{8},\qquad b=\frac{7}{64}.$$ 6. Compute: $$9a+8b=9\cdot\frac{1}{8}+8\cdot\frac{7}{64}$$ $$=\frac{9}{8}+\frac{7}{8}=\frac{16}{8}=2.$$ 7. Hence the correct option is: $$\boxed{\text{A: }2}$$

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Correct answer: A

\left(\frac{1+\cos 2x}{2}\right)^2.$$ So, $$\cos^4 x=\frac{1}{4}\left(1+2\cos 2x+\cos^2 2x\right).$$ Now, $$\cos^2 2x=\frac{1+\cos 4x}{2}.$$ Hence, $$\cos^4 x=\frac{1}{4}\left(1+2\cos 2x+\frac{1+\cos 4x}{2}\right) =\frac{1}{4}\left(\frac{3}{2}+2\cos 2x+\frac{1}{2}\cos 4x\right).$$ Thus, $$\cos^4 x=\frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x.$$ 3. Substitute into the integral: $$I=\int_0^{\pi/3}\left(\frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x\right)dx.$$ Integrate term by term: $$I=\left[\frac{3x}{8}+\frac{1}{2}\cdot\frac{\sin 2x}{2}+\frac{1}{8}\cdot\frac{\sin 4x}{4}\right]_0^{\pi/3}$$ $$=\left[\frac{3x}{8}+\frac{\sin 2x}{4}+\frac{\sin 4x}{32}\right]_0^{\pi/3}.$$ 4. Evaluate at the limits: At $x=\pi/3$, $$\frac{3x}{8}=\frac{3(\pi/3)}{8}=\frac{\pi}{8},$$ $$\sin\left(\frac{2\pi}{3}\right)=\frac{\sqrt{3}}{2},$$ $$\sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}.$$ So, $$I=\frac{\pi}{8}+\frac{1}{4}\cdot\frac{\sqrt{3}}{2}+\frac{1}{32}\cdot\left(-\frac{\sqrt{3}}{2}\right).$$ $$I=\frac{\pi}{8}+\frac{\sqrt{3}}{8}-\frac{\sqrt{3}}{64}.$$ $$I=\frac{\pi}{8}+\frac{8\sqrt{3}-\sqrt{3}}{64}$$ $$I=\frac{\pi}{8}+\frac{7\sqrt{3}}{64}.$$ 5. Compare with $$I=a\pi+b\sqrt{3}.$$ Therefore, $$a=\frac{1}{8},\qquad b=\frac{7}{64}.$$ 6. Compute: $$9a+8b=9\cdot\frac{1}{8}+8\cdot\frac{7}{64}$$ $$=\frac{9}{8}+\frac{7}{8}=\frac{16}{8}=2.$$ 7. Hence the correct option is: $$\boxed{\text{A: }2}$$