JEE MainMathematics

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Definite Integral of Even Powers of Cosine

If integral from 0 to π/3 of cos^4 x dx = aπ + b√(3), where a and b are rational numbers, then 9a + 8b is equal to:

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A2B1C3D3/2

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Correct answer

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

Option analysis

Why each option works or fails

A · 2

The integral is evaluated correctly using trigonometric identities. Correct. Integrating $\cos^{4} x = \frac{1}{8} (3 + 4 \cos 2 x + \cos 4 x)$ from $0$ to $\frac{π}{3}$ yields $a = \frac{1}{8}$ and $b = \frac{7}{64}$ , giving $9 a + 8 b = 2$ .

B · 1

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

C · 3

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

D · 3/2

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

Reviewed route

Solution

StepWorking

01Given

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

02Goal

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

03Approach

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

04Execute

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

05Execute

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

06Execute

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

✓Verify

Check that each component matches standard identities: $\int_{0}^{π / 3} d x = π / 3$ , $\sin (2 π / 3) = \sqrt{3} / 2$ , $\sin (4 π / 3) = - \sqrt{3} / 2$ . Combining yields $a = 1 / 8$ and $b = 7 / 64$ , confirming $9 a + 8 b = 2$ .

Hints that build this answer step by step

Which identity correctly rewrites $\cos^{4} x$ in terms of multiple angles for integration?
$\cos^{4} x = \frac{1}{8} (3 + 4 \cos 2 x + \cos 4 x)$
What is the antiderivative $F (x) = \int \frac{1}{8} (3 + 4 \cos 2 x + \cos 4 x) d x$ ?
$F (x) = \frac{1}{8} (3 x + 2 \sin 2 x + \frac{1}{4} \sin 4 x)$
Evaluating $F (\frac{π}{3}) - F (0)$ , what are the values of $a$ and $b$ in $a π + b \sqrt{3}$ ?
$a = \frac{1}{8}, b = \frac{7}{64}$
With $a = \frac{1}{8}$ and $b = \frac{7}{64}$ , what is $9 a + 8 b$ ?
$9 (\frac{1}{8}) + 8 (\frac{7}{64}) = \frac{9}{8} + \frac{7}{8} = 2$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $\frac{π}{8} + \frac{7 \sqrt{3}}{64}$ , giving $a = \frac{1}{8}$ and $b = \frac{7}{64}$ , so $9 a + 8 b = 2$ .

Why each option works or fails

A · correct: 2 - The integral is evaluated correctly using trigonometric identities. Correct. Integrating $\cos^{4} x = \frac{1}{8} (3 + 4 \cos 2 x + \cos 4 x)$ from $0$ to $\frac{π}{3}$ yields $a = \frac{1}{8}$ and $b = \frac{7}{64}$ , giving $9 a + 8 b = 2$ .
B: 1 - A sign error or arithmetic slip occurred while evaluating $\sin (4 x)$ at $x = \frac{π}{3}$ . Check the value of $\sin (\frac{4 π}{3}) = - \frac{\sqrt{3}}{2}$ carefully when evaluating the antiderivative terms.
C: 3 - Forgetting the division by $4$ when integrating $\cos (4 x)$ , leading to an incorrect coefficient $b$ . Remember that $\int \cos (k x) d x = \frac{1}{k} \sin (k x)$ ; applying the chain rule correctly is essential.
D: 3/2 - Misexpanding ${(\frac{1 + \cos 2 x}{2})}^{2}$ by dropping the constant term or miscalculating the leading coefficient. Ensure $\cos^{4} x = \frac{1}{4} (1 + 2 \cos 2 x + \cos^{2} 2 x) = \frac{3}{8} + \frac{1}{2} \cos 2 x + \frac{1}{8} \cos 4 x$ before integrating.

Step-by-step solution

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

Definite Integral of Even Powers of Cosine

What feels right?

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

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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