If $\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x=\mathrm{a} \pi+\mathrm{b} \sqrt{3}$, where $\mathrm{a}$ and $\mathrm{b}$ are rational numbers, then $9 \mathrm{a}+8 \mathrm{b}$ is equal to :
Solution
<p>To solve the given integral $\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x$, we'll apply a known power-reduction formula that allows us to express even powers of sine and cosine functions in terms of cosine of multiple angles. Specifically for $\cos^4 x$, we can write it in terms of double angles as:</p>
<p>$\cos^4 x = \left(\frac{1 + \cos(2x)}{2}\right)^2$</p>
<p>We can then expand and simplify the integral using this formula. Let's proceed with this:</p>
<p>$$
\int\limits_0^{\frac{\pi}{3}} \cos^4 x \mathrm{~d}x = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1 + \cos(2x)}{2}\right)^2 \mathrm{~d}x
$$</p>
<p>Now, let's expand the integrand and then integrate term by term:</p>
<p>$$
= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \cos^2(2x)\right)\mathrm{~d}x
$$</p>
<p>For the $\cos^2(2x)$ term, we again use the power reduction formula:</p>
<p>$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$</p>
<p>Let's substitute this into the integral and continue:</p>
<p>$$
= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \left(\frac{1 + \cos(4x)}{2}\right)\right)\mathrm{~d}x
$$</p>
<p>Simplify and integrate:</p>
<p>$$
= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{8} + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x
$$</p>
<p>$$
= \int\limits_0^{\frac{\pi}{3}} \left(\frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x
$$</p>
<p>$$
= \left. \left(\frac{3}{8} x + \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x)\right) \right|_0^{\frac{\pi}{3}}
$$</p>
<p>Evaluating this from $0$ to $\frac{\pi}{3}$:</p>
<p>$$
= \left(\frac{3}{8} \cdot \frac{\pi}{3} + \frac{1}{4} \sin\left(2 \cdot \frac{\pi}{3}\right) + \frac{1}{32} \sin\left(4 \cdot \frac{\pi}{3}\right)\right) - \left(\frac{3}{8} \cdot 0 + \frac{1}{4} \sin(0) + \frac{1}{32} \sin(0)\right)
$$</p>
<p>$$
= \frac{\pi}{8} + \frac{1}{4} \sin\left(\frac{2\pi}{3}\right) + \frac{1}{32} \sin\left(\frac{4\pi}{3}\right)
$$</p>
<p>$\sin\left(\frac{2\pi}{3}\right)$ is $\frac{\sqrt{3}}{2}$ and $\sin\left(\frac{4\pi}{3}\right)$ is $-\frac{\sqrt{3}}{2}$:</p>
<p>$$
= \frac{\pi}{8} + \frac{1}{4} \cdot \frac{\sqrt{3}}{2} - \frac{1}{32} \cdot \frac{\sqrt{3}}{2}
$$</p>
<p>$= \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64}$</p>
<p>Now, combining terms we get the final result:</p>
<p>$= \frac{\pi}{8} + \frac{8\sqrt{3} - \sqrt{3}}{64}$</p>
<p>$= \frac{\pi}{8} + \frac{7\sqrt{3}}{64}$</p>
<p>Now, let's match this result to the form $\mathrm{a}\pi + \mathrm{b}\sqrt{3}$ and find $a$ and $b$:</p>
<p>$a = \frac{1}{8}, \quad b = \frac{7}{64}$</p>
<p>Now we find $9a + 8b$:</p>
<p>$9a + 8b = 9 \cdot \frac{1}{8} + 8 \cdot \frac{7}{64}$</p>
<p>$= \frac{9}{8} + \frac{7}{8}$</p>
<p>$= \frac{16}{8}$</p>
<p>$= 2$</p>
<p>Therefore, the value of $9a + 8b$ is 2, which correspond to Option A.</p>
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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