If ∫(0 to π/3) cos⁴ x dx = aπ + b√3, where a and b are rational numbers, then 9a + 8b is equal to :

Q: Why can't we use Wallis' Formula here?

Wallis' Formula is a powerful tool, but it is specifically designed for integrals of the form $\int_{0}^{\pi/2} \sin^n x \cos^m x dx$. Since our upper limit is $\pi/3$, not $\pi/2$, Wallis' Formula does not apply. Always check your limits before applying shortcuts!

Q: Why do we get a negative sign when integrating $\cos 4x$?

Actually, the integral of $\cos 4x$ is $\frac{\sin 4x}{4}$. There is no negative sign here. The negative sign appears only when evaluating the limits, specifically when $\sin(4 \cdot \frac{\pi}{3}) = \sin(\frac{4\pi}{3}) = -\frac{\sqrt{3}}{2}$. Keep a close eye on your quadrant signs!

Q: Is there a simpler way to expand $(1 + \cos 2x)^2$?

You could use the binomial expansion, but $(1 + \cos 2x)^2 = 1 + 2\cos 2x + \cos^2 2x$ is straightforward enough. The key is to be methodical. Don't skip the expansion step, as it's the most common place for arithmetic errors.

Question

If $\int_{0}^{\pi/3} \cos^4 x dx = a\pi + b\sqrt{3}$, where $a$ and $b$ are rational numbers, then $9a + 8b$ is equal to :

Options:

(A) 2

(B) 1

(C) 3

(D) $\frac{3}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the base of a mountain, looking up at the peak. That is how many students feel when they see an integral like $\int_{0}^{\pi/3} \cos^4 x \, dx$. It looks intimidating, almost insurmountable.

In the world of JEE mathematics, no mountain is too high if you have the right tools. Today, we are going to break this problem down by understanding the elegant geometry and algebra hidden beneath the surface.

### The Visualization

Before we touch a single equation, let's visualize what we are doing. We are calculating the area under the curve $y = \cos^4 x$ from $x=0$ to $x=\frac{\pi}{3}$.

Because the power is even, the function is always positive, and the area will be a positive value. Our goal is to find this area and then map it to the form $a\pi + b\sqrt{3}$.

### The Power Reduction Strategy

The biggest mistake students make is trying to integrate $\cos^4 x$ directly. We need to linearize it using a 'divide and conquer' strategy.

We know that $\cos^4 x = (\cos^2 x)^2$. Using the golden identity $\cos^2 x = \frac{1 + \cos 2x}{2}$, we transform our integral into:

$$I = \int_{0}^{\pi/3} \left( \frac{1 + \cos 2x}{2} \right)^2 dx$$

### The Expansion

Now, we expand the square. Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, we obtain:

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

Notice that we have successfully reduced the power from 4 to 2. However, we still have a $\cos^2 2x$ term that requires further reduction.

### The Second Reduction

We apply the same identity again for $\cos^2 2x$. Since the angle doubles, we use $\cos^2 2x = \frac{1 + \cos 4x}{2}$.

Substituting this back into our integral, we get:

$$I = \frac{1}{4} \int_{0}^{\pi/3} \left( 1 + 2\cos 2x + \frac{1 + \cos 4x}{2} \right) dx$$

Simplifying the integrand, we have a constant, a $\cos 2x$ term, and a $\cos 4x$ term. This is now ready for direct integration.

### The Integration

We integrate term by term. The expression simplifies to:

$$I = \frac{1}{4} \int_{0}^{\pi/3} \left( \frac{3}{2} + 2\cos 2x + \frac{1}{2}\cos 4x \right) dx$$

The integral of the constant $\frac{3}{2}$ is $\frac{3x}{2}$. The integral of $2\cos 2x$ is $\sin 2x$, and the integral of $\frac{1}{2}\cos 4x$ is $\frac{\sin 4x}{8}$.

### Final Calculation

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

$$I = \frac{1}{4} \left[ \frac{3}{2}x + \sin 2x + \frac{\sin 4x}{8} \right]_{0}^{\pi/3}$$

Substituting the limits:

$$I = \frac{1}{4} \left[ \frac{3}{2} \left( \frac{\pi}{3} \right) + \sin \left( \frac{2\pi}{3} \right) + \frac{\sin(4\pi/3)}{8} \right] = \frac{\pi}{8} + \frac{7\sqrt{3}}{64}$$

Comparing this with $a\pi + b\sqrt{3}$, we identify $a = \frac{1}{8}$ and $b = \frac{7}{64}$. Finally, calculating $9a + 8b$ gives us:

$$9 \left( \frac{1}{8} \right) + 8 \left( \frac{7}{64} \right) = \frac{9}{8} + \frac{7}{8} = \mathbf{2}$$

Animated Solution for Mathematics - Definite Integration: If $\int_{0}^{π /3} cos^{4} x d x = aπ + b 3$ , where $a$ and $b$ are rational numbers, then $9 a + 8 b$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Integral

Power Reduction - Step 1

Expanding the Square

Power Reduction - Step 2

Simplifying the Integrand

Performing Integration

Applying Upper Limit

Applying Lower Limit

Final Simplification

Finding a and b

Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Visualization

The Power Reduction Strategy

The Expansion

The Second Reduction

The Integration

Final Calculation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫0π/3​cos4xdx=aπ+b3​, where a and b are rational numbers, then 9a+8b is equal to :