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Integral Calculus: JEE Main Mathematics Question with Solution

\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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2

1

3

\frac{3}{2}

Step-by-step solutionView

Correct answer

Using power-reduction identities, the integral evaluates to $\frac{\pi}{8} + \frac{7\sqrt{3}}{64}$ , giving $a = \frac{1}{8}$ and $b = \frac{7}{64}$ , so $9a + 8b = 2$ .

Option analysis

Why each option works or fails

A ·

2

None. The integral is evaluated correctly using trigonometric identities. Correct. Integrating $\cos^4 x = \frac{1}{8}(3 + 4\cos 2x + \cos 4x)$ from $0$ to $\frac{\pi}{3}$ yields $a = \frac{1}{8}$ and $b = \frac{7}{64}$ , giving $9a + 8b = 2$ .

B ·

1

A sign error or arithmetic slip occurred while evaluating $\sin(4x)$ at $x = \frac{\pi}{3}$ . Check the value of $\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}$ carefully when evaluating the antiderivative terms.

C ·

3

Forgetting the division by $4$ when integrating $\cos(4x)$ , leading to an incorrect coefficient $b$ . Remember that $\int \cos(kx)\,dx = \frac{1}{k}\sin(kx)$ ; applying the chain rule correctly is essential.

D ·

\frac{3}{2}

Misexpanding $\left(\frac{1+\cos 2x}{2}\right)^2$ by dropping the constant term or miscalculating the leading coefficient. Ensure $\cos^4 x = \frac{1}{4}(1 + 2\cos 2x + \cos^2 2x) = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$ before integrating.

Reviewed route

Solution

StepWorking

01given

$I = \int_{0}^{\frac{\pi}{3}} \cos^{4} x \, dx = a\pi + b\sqrt{3}$ , where $a, b \in \mathbb{Q}$ .

02goal

Evaluate the definite integral to find rational constants $a$ and $b$ , and compute $9a + 8b$ .

03approach

Use the power-reduction formula $\cos^2 x = \frac{1 + \cos 2x}{2}$ , expand $\cos^4 x$ , reduce $\cos^2 2x$ using $\frac{1 + \cos 4x}{2}$ , and integrate term-by-term from $0$ to $\frac{\pi}{3}$ .

04execute

Express the integrand using multiple angles: $\cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1}{4}\left(1 + 2\cos 2x + \cos^2 2x\right) = \frac{1}{4}\left(1 + 2\cos 2x + \frac{1 + \cos 4x}{2}\right) = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$

05execute

Integrate term-by-term: $\int_{0}^{\pi/3} \left(\frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x\right) dx = \left[ \frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x \right]_{0}^{\pi/3}$ Evaluate at upper limit $x = \frac{\pi}{3}$ : $\frac{3}{8}\left(\frac{\pi}{3}\right) + \frac{1}{4}\sin\left(\frac{2\pi}{3}\right) + \frac{1}{32}\sin\left(\frac{4\pi}{3}\right) = \frac{\pi}{8} + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{32}\left(-\frac{\sqrt{3}}{2}\right)$ $= \frac{\pi}{8} + \sqrt{3}\left(\frac{1}{8} - \frac{1}{64}\right) = \frac{\pi}{8} + \frac{7\sqrt{3}}{64}$

06execute

Identify $a$ and $b$ : $a = \frac{1}{8}, \quad b = \frac{7}{64}$ Compute the linear combination: $9a + 8b = 9\left(\frac{1}{8}\right) + 8\left(\frac{7}{64}\right) = \frac{9}{8} + \frac{7}{8} = \frac{16}{8} = 2$

✓verify

Check that each component matches standard identities: $\int_0^{\pi/3} dx = \pi/3$ , $\sin(2\pi/3) = \sqrt{3}/2$ , $\sin(4\pi/3) = -\sqrt{3}/2$ . Combining yields $a = 1/8$ and $b = 7/64$ , confirming $9a + 8b = 2$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why not use integration by parts or reduction formula directly?

Double-angle power reduction expands the degree 4 polynomial in cosine directly into linear harmonic terms, which allows immediate antiderivative evaluation without recursion steps.

Answer

Using power-reduction identities, the integral evaluates to $\frac{\pi}{8} + \frac{7\sqrt{3}}{64}$ , giving $a = \frac{1}{8}$ and $b = \frac{7}{64}$ , so $9a + 8b = 2$ .

Why each option works or fails

A · correct: $2$ - None. The integral is evaluated correctly using trigonometric identities. Correct. Integrating $\cos^4 x = \frac{1}{8}(3 + 4\cos 2x + \cos 4x)$ from $0$ to $\frac{\pi}{3}$ yields $a = \frac{1}{8}$ and $b = \frac{7}{64}$ , giving $9a + 8b = 2$ .
B: $1$ - A sign error or arithmetic slip occurred while evaluating $\sin(4x)$ at $x = \frac{\pi}{3}$ . Check the value of $\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}$ carefully when evaluating the antiderivative terms.
C: $3$ - Forgetting the division by $4$ when integrating $\cos(4x)$ , leading to an incorrect coefficient $b$ . Remember that $\int \cos(kx)\,dx = \frac{1}{k}\sin(kx)$ ; applying the chain rule correctly is essential.
D: $9a + 8b = 2$ - Misexpanding $\left(\frac{1+\cos 2x}{2}\right)^2$ by dropping the constant term or miscalculating the leading coefficient. Ensure $\cos^4 x = \frac{1}{4}(1 + 2\cos 2x + \cos^2 2x) = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$ before integrating.

Step-by-step solution

given: $I = \int_{0}^{\frac{\pi}{3}} \cos^{4} x \, dx = a\pi + b\sqrt{3}$ , where $a, b \in \mathbb{Q}$ .
goal: Evaluate the definite integral to find rational constants $a$ and $3$ , and compute $9a + 8b$ .
approach: Use the power-reduction formula $\cos^2 x = \frac{1 + \cos 2x}{2}$ , expand $\cos^4 x$ , reduce $\cos^2 2x$ using $\frac{1 + \cos 4x}{2}$ , and integrate term-by-term from $0$ to $\frac{\pi}{3}$ .
execute: Express the integrand using multiple angles: $\cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1}{4}\left(1 + 2\cos 2x + \cos^2 2x\right) = \frac{1}{4}\left(1 + 2\cos 2x + \frac{1 + \cos 4x}{2}\right) = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$
execute: Integrate term-by-term: $\int_{0}^{\pi/3} \left(\frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x\right) dx = \left[ \frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x \right]_{0}^{\pi/3}$ Evaluate at upper limit $x = \frac{\pi}{3}$ : $\frac{3}{8}\left(\frac{\pi}{3}\right) + \frac{1}{4}\sin\left(\frac{2\pi}{3}\right) + \frac{1}{32}\sin\left(\frac{4\pi}{3}\right) = \frac{\pi}{8} + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{32}\left(-\frac{\sqrt{3}}{2}\right)$ $= \frac{\pi}{8} + \sqrt{3}\left(\frac{1}{8} - \frac{1}{64}\right) = \frac{\pi}{8} + \frac{7\sqrt{3}}{64}$
execute: Identify $a$ and $b$ : $a = \frac{1}{8}, \quad b = \frac{7}{64}$ Compute the linear combination: $9a + 8b = 9\left(\frac{1}{8}\right) + 8\left(\frac{7}{64}\right) = \frac{9}{8} + \frac{7}{8} = \frac{16}{8} = 2$
verify: Check that each component matches standard identities: $\int_0^{\pi/3} dx = \pi/3$ , $\sin(2\pi/3) = \sqrt{3}/2$ , $\sin(4\pi/3) = -\sqrt{3}/2$ . Combining yields $a = 1/8$ and $b = 7/64$ , confirming $9a + 8b = 2$ .

Integral Calculus: JEE Main Mathematics Question with Solution

Using power-reduction identities, the integral evaluates to π8+7364\frac{\pi}{8} + \frac{7\sqrt{3}}{64}8π​+6473​​, giving a=18a = \frac{1}{8}a=81​ and b=764b = \frac{7}{64}b=647​, so 9a+8b=29a + 8b = 29a+8b=2.

01given

02goal

03approach

04execute

05execute

06execute

✓verify

Students also ask

Using power-reduction identities, the integral evaluates to $\frac{\pi}{8} + \frac{7\sqrt{3}}{64}$ , giving $a = \frac{1}{8}$ and $b = \frac{7}{64}$ , so $9a + 8b = 2$ .