JEE MainMathematics

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Definite Integral of Even Trigonometric Powers via Periodicity

Q: Why is the period of sin^2(2x) equal to π/2?

The period of sin(x) is 2π, so the period of sin(2x) is π. Squaring a sinusoidal function halves its fundamental period because sin^2(θ) = (1 - cos(2θ))/2, which has period π. Therefore, sin^2(2x) = (1 - cos(4x))/2, having period 2π/4 = π/2.

The value of integral from 0 to 20π of (sin^4 x + cos^4 x)dx is equal to:

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A15π/2B25πC15πD25π/2

Step-by-step solutionView

Correct answer

Using trigonometric identities and the periodicity of $\sin^{4} x + \cos^{4} x$ , the integral evaluates to $15 π$ .

Option analysis

Why each option works or fails

A · 15π/2

Believing that the interval $[0, 20 π]$ contains only 10 periods of $π$ , or inadvertently halving the result during the periodic reduction. The fundamental period of $\sin^{4} x + \cos^{4} x$ is $\frac{π}{2}$ , meaning the interval $[0, 20 π]$ contains 40 periods of length $\frac{π}{2}$ , or 20 periods of length $π$ .

B · 25π

Incorrectly rewriting $\sin^{4} x + \cos^{4} x$ as $1 + 2 \sin^{2} x \cos^{2} x$ instead of $1 - 2 \sin^{2} x \cos^{2} x$ , yielding an average value of $\frac{5}{4}$ instead of $\frac{3}{4}$ . Apply the identity $a^{2} + b^{2} = (a + b)^{2} - 2 a b$ ; hence $\sin^{4} x + \cos^{4} x = (\sin^{2} x + \cos^{2} x)^{2} - 2 \sin^{2} x \cos^{2} x = 1 - \frac{1}{2} \sin^{2} 2 x$ .

C · 15π

This is the correct answer. Rewriting the integrand as $1 - \frac{1}{2} \sin^{2} (2 x) = \frac{3}{4} + \frac{1}{4} \cos (4 x)$ , the average value over any full cycle is $\frac{3}{4}$ . Multiplying by the interval length $20 π$ gives $20 π \times \frac{3}{4} = 15 π$ .

D · 25π/2

Confusing the sign in the reduction formula and dividing by 2 incorrectly, leading to an extra factor of $\frac{5}{8}$ or $\frac{5}{4}$ . Ensure signs are tracked carefully when expanding $(\sin^{2} x + \cos^{2} x)^{2} - 2 \sin^{2} x \cos^{2} x$ and integrating over the full length of $20 π$ .

Reviewed route

Solution

StepWorking

01Given

The integral to evaluate is $I = \int_{0}^{20 π} (\sin^{4} x + \cos^{4} x) d x$ .

02Goal

Evaluate the exact value of the integral $I$ .

03Approach

Simplify the integrand using the identity $\sin^{4} x + \cos^{4} x = 1 - 2 \sin^{2} x \cos^{2} x = 1 - \frac{1}{2} \sin^{2} (2 x)$ , then use the periodicity of $\sin^{2} (2 x)$ to evaluate the integral over one fundamental period.

04Execute

Rewrite the integrand: $\sin^{4} x + \cos^{4} x = (\sin^{2} x + \cos^{2} x)^{2} - 2 \sin^{2} x \cos^{2} x = 1 - \frac{1}{2} \sin^{2} (2 x)$ So, $I = \int_{0}^{20 π} 1 d x - \frac{1}{2} \int_{0}^{20 π} \sin^{2} (2 x) d x = 20 π - \frac{1}{2} I_{1}$

05Execute

The function $\sin^{2} (2 x)$ is periodic with period $T = \frac{π}{2}$ . The interval $[0, 20 π]$ contains $\frac{20 π}{π / 2} = 40$ periods. $I_{1} = 40 \int_{0}^{π / 2} \sin^{2} (2 x) d x = 40 \int_{0}^{π / 2} \frac{1 - \cos (4 x)}{2} d x = 40 {[\frac{x}{2} - \frac{\sin (4 x)}{8}]}_{0}^{π / 2} = 40 (\frac{π}{4}) = 10 π$

06Execute

Substitute $I_{1} = 10 π$ back into $I$ : $I = 20 π - \frac{1}{2} (10 π) = 20 π - 5 π = 15 π$

✓Verify

The average value of $\sin^{4} x + \cos^{4} x$ over any full period is $1 - \frac{1}{2} (\frac{1}{2}) = \frac{3}{4}$ . The length of the interval is $20 π$ . Thus, $I = average \times length = \frac{3}{4} \times 20 π = 15 π$ . This verifies the result.

✓ 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: 8 September 2026

Quick checks

Students also ask

Why is the period of

\sin^{2} (2 x)

equal to

π / 2

The period of $\sin (x)$ is $2 π$ , so the period of $\sin (2 x)$ is $π$ . Squaring a sinusoidal function halves its fundamental period because $\sin^{2} (θ) = \frac{1 - \cos (2 θ)}{2}$ , which has period $π$ . Therefore, $\sin^{2} (2 x) = \frac{1 - \cos (4 x)}{2}$ , having period $\frac{2 π}{4} = \frac{π}{2}$ .

Answer

Using trigonometric identities and the periodicity of $\sin^{4} x + \cos^{4} x$ , the integral evaluates to $15 π$ .

Why each option works or fails

A: 15π/2 - Believing that the interval $[0, 20 π]$ contains only 10 periods of $π$ , or inadvertently halving the result during the periodic reduction. The fundamental period of $\sin^{4} x + \cos^{4} x$ is $\frac{π}{2}$ , meaning the interval $[0, 20 π]$ contains 40 periods of length $\frac{π}{2}$ , or 20 periods of length $π$ .
B: 25π - Incorrectly rewriting $\sin^{4} x + \cos^{4} x$ as $1 + 2 \sin^{2} x \cos^{2} x$ instead of $1 - 2 \sin^{2} x \cos^{2} x$ , yielding an average value of $\frac{5}{4}$ instead of $\frac{3}{4}$ . Apply the identity $a^{2} + b^{2} = (a + b)^{2} - 2 a b$ ; hence $\sin^{4} x + \cos^{4} x = (\sin^{2} x + \cos^{2} x)^{2} - 2 \sin^{2} x \cos^{2} x = 1 - \frac{1}{2} \sin^{2} 2 x$ .
C · correct: 15π - This is the correct answer. Rewriting the integrand as $1 - \frac{1}{2} \sin^{2} (2 x) = \frac{3}{4} + \frac{1}{4} \cos (4 x)$ , the average value over any full cycle is $\frac{3}{4}$ . Multiplying by the interval length $20 π$ gives $20 π \times \frac{3}{4} = 15 π$ .
D: 25π/2 - Confusing the sign in the reduction formula and dividing by 2 incorrectly, leading to an extra factor of $\frac{5}{8}$ or $\frac{5}{4}$ . Ensure signs are tracked carefully when expanding $(\sin^{2} x + \cos^{2} x)^{2} - 2 \sin^{2} x \cos^{2} x$ and integrating over the full length of $20 π$ .

Step-by-step solution

Given: The integral to evaluate is $I = \int_{0}^{20 π} (\sin^{4} x + \cos^{4} x) d x$ .
Goal: Evaluate the exact value of the integral $I$ .
Approach: Simplify the integrand using the identity $\sin^{4} x + \cos^{4} x = 1 - 2 \sin^{2} x \cos^{2} x = 1 - \frac{1}{2} \sin^{2} (2 x)$ , then use the periodicity of $\sin^{2} (2 x)$ to evaluate the integral over one fundamental period.
Execute: Rewrite the integrand: $\sin^{4} x + \cos^{4} x = (\sin^{2} x + \cos^{2} x)^{2} - 2 \sin^{2} x \cos^{2} x = 1 - \frac{1}{2} \sin^{2} (2 x)$ So, $I = \int_{0}^{20 π} 1 d x - \frac{1}{2} \int_{0}^{20 π} \sin^{2} (2 x) d x = 20 π - \frac{1}{2} I_{1}$
Execute: The function $\sin^{2} (2 x)$ is periodic with period $T = \frac{π}{2}$ . The interval $[0, 20 π]$ contains $\frac{20 π}{π / 2} = 40$ periods. $I_{1} = 40 \int_{0}^{π / 2} \sin^{2} (2 x) d x = 40 \int_{0}^{π / 2} \frac{1 - \cos (4 x)}{2} d x = 40 {[\frac{x}{2} - \frac{\sin (4 x)}{8}]}_{0}^{π / 2} = 40 (\frac{π}{4}) = 10 π$
Execute: Substitute $I_{1} = 10 π$ back into $I$ : $I = 20 π - \frac{1}{2} (10 π) = 20 π - 5 π = 15 π$
Verify: The average value of $\sin^{4} x + \cos^{4} x$ over any full period is $1 - \frac{1}{2} (\frac{1}{2}) = \frac{3}{4}$ . The length of the interval is $20 π$ . Thus, $I = average \times length = \frac{3}{4} \times 20 π = 15 π$ . This verifies the result.

Shortcut:

When to use it: When integrating periodic functions over an integer number of fundamental periods.

Given: $I = \int_{0}^{20 π} f (x) d x$ where $f (x) = \sin^{4} x + \cos^{4} x = \frac{3}{4} + \frac{1}{4} \cos (4 x)$ .

Goal: Quickly evaluate the integral across an integer number of full periods.

Approach: The oscillating term $\cos (4 x)$ integrates to 0 over any integer number of periods (length $20 π$ is 40 full periods of $\cos (4 x)$ ). Only the constant term contributes.

Execute: $I = \int_{0}^{20 π} (\frac{3}{4} + \frac{1}{4} \cos (4 x)) d x = \frac{3}{4} \times 20 π + 0 = 15 π$

Verify: Matches option C.

Definite Integral of Even Trigonometric Powers via Periodicity

What feels right?

Using trigonometric identities and the periodicity of $\sin^{4} x + \cos^{4} x$ , the integral evaluates to $15 π$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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