If int_0^{frac{pi}{3}} cos^4 x , dx = api + b | vedclass

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(A) We evaluate the integral $I = \int_0^{\pi/3} \cos^4 x \, dx$.Using the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$,we have $\cos^4 x = \left(\frac

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$2$

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(A) We evaluate the integral $I = \int_0^{\pi/3} \cos^4 x \, dx$.Using the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$,we have $\cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1}{4}(1 + 2\cos 2x + \cos^2 2x)$.Substituting $\cos^2 2x = \frac{1 + \cos 4x}{2}$,we get $\cos^4 x = \frac{1}{4} + \frac{1}{2}\cos 2x + \frac{1}{8} + \frac{1}{8}\cos 4x = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$.Now,integrate term by term:$I = \int_0^{\pi/3} \left(\frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x\right) dx$$I = \left[ \frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x \right]_0^{\pi/3}$$I = \left( \frac{3}{8} \cdot \frac{\pi}{3} + \frac{1}{4}\sin\frac{2\pi}{3} + \frac{1}{32}\sin\frac{4\pi}{3} \right) - (0)$$I = \frac{\pi}{8} + \frac{1}{4} \cdot \frac{\sqrt{3}}{2} + \frac{1}{32} \cdot \left(-\frac{\sqrt{3}}{2}\right)$$I = \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64} = \frac{\pi}{8} + \frac{8\sqrt{3} - \sqrt{3}}{64} = \frac{\pi}{8} + \frac{7\sqrt{3}}{64}$.Comparing with $a\pi + b\sqrt{3}$,we get $a = \frac{1}{8}$ and $b = \frac{7}{64}$.Thus,$9a + 8b = 9(\frac{1}{8}) + 8(\frac{7}{64}) = \frac{9}{8} + \frac{7}{8} = \frac{16}{8} = 2$.

If $\int_0^{\frac{\pi}{3}} \cos^4 x \, dx = a\pi + b\sqrt{3}$,where $a$ and $b$ are rational numbers,then $9a + 8b$ is equal to:

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