QJEEM24Mad05a: JEE Main 2024 Mathematics - Class 12 JEEM24Mad05a Solution: If , where and are rational numbers, then is equal to :

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:

$\cos^4 x = \left(\frac{1 + \cos(2x)}{2}\right)^2$

We can then expand and simplify the integral using this formula. Let's proceed with this:

$\int\limits_0^{\frac{\pi}{3}} \cos^4 x \mathrm{~d}x = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1 + \cos(2x)}{2}\right)^2 \mathrm{~d}x$

Now, let's expand the integrand and then integrate term by term:

$= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \cos^2(2x)\right)\mathrm{~d}x$

For the $\cos^2(2x)$ term, we again use the power reduction formula:

$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$

Let's substitute this into the integral and continue:

$= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \left(\frac{1 + \cos(4x)}{2}\right)\right)\mathrm{~d}x$

Simplify and integrate:

$= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{8} + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x$

$= \int\limits_0^{\frac{\pi}{3}} \left(\frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x$

$= \left. \left(\frac{3}{8} x + \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x)\right) \right|_0^{\frac{\pi}{3}}$

Evaluating this from $0$ to $\frac{\pi}{3}$ :

$= \left(\frac{3}{8} \cdot \frac{\pi}{3} + \frac{1}{4} \sin\left(2 \cdot \frac{\pi}{3}\right) + \frac{1}{32} \sin\left(4 \cdot \frac{\pi}{3}\right)\right) - \left(\frac{3}{8} \cdot 0 + \frac{1}{4} \sin(0) + \frac{1}{32} \sin(0)\right)$

$= \frac{\pi}{8} + \frac{1}{4} \sin\left(\frac{2\pi}{3}\right) + \frac{1}{32} \sin\left(\frac{4\pi}{3}\right)$

$\sin\left(\frac{2\pi}{3}\right)$ is $\frac{\sqrt{3}}{2}$ and $\sin\left(\frac{4\pi}{3}\right)$ is $-\frac{\sqrt{3}}{2}$ :

$= \frac{\pi}{8} + \frac{1}{4} \cdot \frac{\sqrt{3}}{2} - \frac{1}{32} \cdot \frac{\sqrt{3}}{2}$

$= \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64}$

Now, combining terms we get the final result:

$= \frac{\pi}{8} + \frac{8\sqrt{3} - \sqrt{3}}{64}$

$= \frac{\pi}{8} + \frac{7\sqrt{3}}{64}$

Now, let's match this result to the form $\mathrm{a}\pi + \mathrm{b}\sqrt{3}$ and find $a$ and $b$ :

$a = \frac{1}{8}, \quad b = \frac{7}{64}$

Now we find $9a + 8b$ :

$9a + 8b = 9 \cdot \frac{1}{8} + 8 \cdot \frac{7}{64}$

$= \frac{9}{8} + \frac{7}{8}$

$= \frac{16}{8}$

$= 2$

Therefore, the value of $9a + 8b$ is 2, which correspond to Option A.