JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectSingle correctpyq

Integral Calculus: JEE Main Mathematics Question with Solution

Q: Why is the period of equal to ?

The period of is , so the period of is . Squaring a sinusoidal function halves its fundamental period because , which has period . Therefore, , having period .

The value of

\int_{0}^{20\pi} (\sin^4 x + \cos^4 x)dx

is equal to :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{15\pi}{2}

25\pi

15\pi

\frac{25\pi}{2}

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A ·

\frac{15\pi}{2}

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

B ·

25\pi

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 $a^2 + b^2 = (a+b)^2 - 2ab$ ; hence $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x$ .

C ·

15\pi

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

D ·

\frac{25\pi}{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 $20\pi$ .

Reviewed route

Solution

StepWorking

01given

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

02goal

Evaluate the exact value of the integral $I$ .

03approach

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

04execute

Rewrite the integrand: $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2(2x)$ So, $I = \int_0^{20\pi} 1 \, dx - \frac{1}{2} \int_0^{20\pi} \sin^2(2x) \, dx = 20\pi - \frac{1}{2} I_1$

05execute

The function $\sin^2(2x)$ is periodic with period $T = \frac{\pi}{2}$ . The interval $[0, 20\pi]$ contains $\frac{20\pi}{\pi/2} = 40$ periods. $I_1 = 40 \int_0^{\pi/2} \sin^2(2x) \, dx = 40 \int_0^{\pi/2} \frac{1 - \cos(4x)}{2} \, dx = 40 \left[ \frac{x}{2} - \frac{\sin(4x)}{8} \right]_0^{\pi/2} = 40 \left( \frac{\pi}{4} \right) = 10\pi$

06execute

Substitute $I_1 = 10\pi$ back into $I$ : $I = 20\pi - \frac{1}{2}(10\pi) = 20\pi - 5\pi = 15\pi$

✓verify

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

✓ 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: 8 September 2026

Quick checks

Students also ask

Why is the period of

\sin^2(2x)

equal to

\pi/2

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

Answer

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

Why each option works or fails

A: $\frac{15\pi}{2}$ - Believing that the interval $[0, 20\pi]$ contains only 10 periods of $\pi$ , or inadvertently halving the result during the periodic reduction. The fundamental period of $\sin^4 x + \cos^4 x$ is $\frac{\pi}{2}$ , meaning the interval $[0, 20\pi]$ contains 40 periods of length $\frac{\pi}{2}$ , or 20 periods of length $\pi$ .
B: $25\pi$ - 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 $\frac{\pi}{2}$ ; hence $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x$ .
C · correct: $15\pi$ - This is the correct answer. Rewriting the integrand as $1 - \frac{1}{2}\sin^2(2x) = \frac{3}{4} + \frac{1}{4}\cos(4x)$ , the average value over any full cycle is $\frac{3}{4}$ . Multiplying by the interval length $20\pi$ gives $20\pi \times \frac{3}{4} = 15\pi$ .
D: $\frac{25\pi}{2}$ - 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 $20\pi$ .

Step-by-step solution

given: The integral to evaluate is $I = \int_{0}^{20\pi} (\sin^4 x + \cos^4 x) \, dx$ .
goal: Evaluate the exact value of the integral $I$ .
approach: Simplify the integrand using the identity $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2(2x)$ , then use the periodicity of $\sin^2(2x)$ to evaluate the integral over one fundamental period.
execute: Rewrite the integrand: $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2(2x)$ So, $I = \int_0^{20\pi} 1 \, dx - \frac{1}{2} \int_0^{20\pi} \sin^2(2x) \, dx = 20\pi - \frac{1}{2} I_1$
execute: The function $\sin^2(2x)$ is periodic with period $T = \frac{\pi}{2}$ . The interval $[0, 20\pi]$ contains $\frac{20\pi}{\pi/2} = 40$ periods. $I_1 = 40 \int_0^{\pi/2} \sin^2(2x) \, dx = 40 \int_0^{\pi/2} \frac{1 - \cos(4x)}{2} \, dx = 40 \left[ \frac{x}{2} - \frac{\sin(4x)}{8} \right]_0^{\pi/2} = 40 \left( \frac{\pi}{4} \right) = 10\pi$
execute: Substitute $I_1 = 10\pi$ back into $I$ : $I = 20\pi - \frac{1}{2}(10\pi) = 20\pi - 5\pi = 15\pi$
verify: The average value of $\sin^4 x + \cos^4 x$ over any full period is $1 - \frac{1}{2}\left(\frac{1}{2}\right) = \frac{3}{4}$ . The length of the interval is $20\pi$ . Thus, $I = \text{average} \times \text{length} = \frac{3}{4} \times 20\pi = 15\pi$ . This verifies the result.

Shortcut:

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

given: $I = \int_0^{20\pi} f(x) \, dx$ where $f(x) = \sin^4 x + \cos^4 x = \frac{3}{4} + \frac{1}{4}\cos(4x)$ .

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

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

execute: $I = \int_0^{20\pi} \left( \frac{3}{4} + \frac{1}{4}\cos(4x) \right) dx = \frac{3}{4} \times 20\pi + 0 = 15\pi$

verify: Matches option (2).

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

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

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

We think you should solve this next

Let…

If the area of the region…

If $\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function satisfying…

Students also ask

Integral Calculus: JEE Main Mathematics Question with Solution

Using trigonometric identities and the periodicity of sin⁡4x+cos⁡4x\sin^4 x + \cos^4 xsin4x+cos4x, the integral evaluates to 15π15\pi15π.

01given

02goal

03approach

04execute

05execute

06execute

✓verify

Let…

If the area of the region…

If f:R→R\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}f:R→R be a continuous function satisfying…

Students also ask

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

If $\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function satisfying…