JEE Main 2024 — Definite Integration Question with Solution

To solve the given integral $\underset{0}{\overset{\frac{\pi }{3}}{\int }}{\cos }^{4}x\mathrm{d}x$, 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:

$$\underset{0}{\overset{\frac{\pi }{3}}{\int }}{\cos }^{4}x\mathrm{d}x=\underset{0}{\overset{\frac{\pi }{3}}{\int }}{\left(\frac{1+\cos (2x)}{2}\right)}^2\mathrm{d}x$$

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

$$=\underset{0}{\overset{\frac{\pi }{3}}{\int }}\left(\frac{1}{4}+\frac{1}{2}\cos (2x)+\frac{1}{4}{\cos }^2(2x)\right)\mathrm{d}x$$

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:

$$=\underset{0}{\overset{\frac{\pi }{3}}{\int }}\left(\frac{1}{4}+\frac{1}{2}\cos (2x)+\frac{1}{4}\left(\frac{1+\cos (4x)}{2}\right)\right)\mathrm{d}x$$

Simplify and integrate:

$$=\underset{0}{\overset{\frac{\pi }{3}}{\int }}\left(\frac{1}{4}+\frac{1}{2}\cos (2x)+\frac{1}{8}+\frac{1}{8}\cos (4x)\right)\mathrm{d}x$$

$$=\underset{0}{\overset{\frac{\pi }{3}}{\int }}\left(\frac{3}{8}+\frac{1}{2}\cos (2x)+\frac{1}{8}\cos (4x)\right)\mathrm{d}x$$

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

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 $b$:

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

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.