Why do all the cosine harmonic terms vanish when integrated from to ?

Because for all positive integers .

JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectNumericalpyq

Integral Calculus: JEE Main Mathematics Question with Solution

\int_{0}^{\pi} \frac{5^{\cos x}(1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x)dx}{1 + 5^{\cos x}} = \frac{k\pi}{16}

, then

k

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.

Type the value - units or words beside it are fine.

Step-by-step solutionView

Correct answer

The value of k is 13.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

The integral is $I = \int_{0}^{\pi} \frac{5^{\cos x}(1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x)}{1 + 5^{\cos x}}\,dx = \frac{k\pi}{16}$ .

02approach

Notice that under $x \to \pi - x$ , $\cos(\pi - x) = -\cos x$ and $\cos 3(\pi - x) = -\cos 3x$ . The trigonometric numerator is invariant because every term involves an even product of cosines: $(-1)(-1) = 1$ , $(-1)^2 = 1$ , and $(-1)^3(-1) = 1$ . Using King's property $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ eliminates the exponential factor $\frac{5^{\cos x}}{1+5^{\cos x}}$ .

03execute

Applying King's rule: $I = \int_{0}^{\pi} \frac{5^{-\cos x}(1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x)}{1 + 5^{-\cos x}}\,dx$ . Adding both expressions for $I$ : $2I = \int_{0}^{\pi} (1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x)\,dx$ . Since the integrand $f(x)$ satisfies $f(\pi - x) = f(x)$ : $2I = 2\int_{0}^{\pi/2} (1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x)\,dx$ $\implies I = \int_{0}^{\pi/2} (1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x)\,dx$ .

04execute

Factor the integrand: $1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x = (1 + \cos^2 x)(1 + \cos x\cos 3x)$ . Alternatively, evaluate directly using trigonometric identities: $\cos x\cos 3x = \frac{1}{2}(\cos 4x + \cos 2x)$ , $\cos^2 x = \frac{1 + \cos 2x}{2}$ , $\cos^3 x\cos 3x = \left(\frac{\cos 3x + 3\cos x}{4}\right)\cos 3x = \frac{1}{4}\cos^2 3x + \frac{3}{4}\cos x\cos 3x = \frac{1+\cos 6x}{8} + \frac{3}{8}(\cos 4x + \cos 2x)$ . Integrating from $0$ to $\frac{\pi}{2}$ , all oscillating terms $\cos(2nx)$ integrate to zero over $[0, \pi/2]$ when $n$ is integer except at the boundaries if any, but specifically: $\int_0^{\pi/2} \cos 2x\,dx = 0$ , $\int_0^{\pi/2} \cos 4x\,dx = 0$ , $\int_0^{\pi/2} \cos 6x\,dx = 0$ . Therefore, the integral only picks up the constant terms: $\text{Constant term} = 1 + 0 + \frac{1}{2} + \frac{1}{8} = 1 + \frac{1}{2} + \frac{1}{8} = \frac{13}{8}$ . Thus, $I = \frac{13}{8} \times \frac{\pi}{2} = \frac{13\pi}{16}$ .

✓verify

Comparing with $I = \frac{k\pi}{16}$ , we get $k = 13$ .

✓ Source and academic review

Question type: Numerical
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 do all the cosine harmonic terms

\cos(2nx)

vanish when integrated from

0

\pi/2

Because $\int_0^{\pi/2} \cos(2nx)dx = \left[\frac{\sin(2nx)}{2n}\right]_0^{\pi/2} = \frac{\sin(n\pi) - 0}{2n} = 0$ for all positive integers $n$ .

Answer

The value of k is 13.

Why each option works or fails

Step-by-step solution

given: The integral is $I = \int_{0}^{\pi} \frac{5^{\cos x}(1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x)}{1 + 5^{\cos x}}\,dx = \frac{k\pi}{16}$ .
approach: Notice that under $x \to \pi - x$ , $\cos(\pi - x) = -\cos x$ and $\cos 3(\pi - x) = -\cos 3x$ . The trigonometric numerator is invariant because every term involves an even product of cosines: $(-1)(-1) = 1$ , $(-1)^2 = 1$ , and $(-1)^3(-1) = 1$ . Using King's property $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ eliminates the exponential factor $\frac{5^{\cos x}}{1+5^{\cos x}}$ .
execute: Applying King's rule: $I = \int_{0}^{\pi} \frac{5^{-\cos x}(1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x)}{1 + 5^{-\cos x}}\,dx$ . Adding both expressions for $I$ : $2I = \int_{0}^{\pi} (1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x)\,dx$ . Since the integrand $f(x)$ satisfies $f(\pi - x) = f(x)$ : $2I = 2\int_{0}^{\pi/2} (1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x)\,dx$ $\implies I = \int_{0}^{\pi/2} (1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x)\,dx$ .
execute: Factor the integrand: $1 + \cos x\cos 3x + \cos^2 x + \cos^3 x\cos 3x = (1 + \cos^2 x)(1 + \cos x\cos 3x)$ . Alternatively, evaluate directly using trigonometric identities: $\cos x\cos 3x = \frac{1}{2}(\cos 4x + \cos 2x)$ , $\cos^2 x = \frac{1 + \cos 2x}{2}$ , $\cos^3 x\cos 3x = \left(\frac{\cos 3x + 3\cos x}{4}\right)\cos 3x = \frac{1}{4}\cos^2 3x + \frac{3}{4}\cos x\cos 3x = \frac{1+\cos 6x}{8} + \frac{3}{8}(\cos 4x + \cos 2x)$ . Integrating from $0$ to $\frac{\pi}{2}$ , all oscillating terms $\cos(2nx)$ integrate to zero over $[0, \pi/2]$ when $n$ is integer except at the boundaries if any, but specifically: $\int_0^{\pi/2} \cos 2x\,dx = 0$ , $\int_0^{\pi/2} \cos 4x\,dx = 0$ , $\int_0^{\pi/2} \cos 6x\,dx = 0$ . Therefore, the integral only picks up the constant terms: $\text{Constant term} = 1 + 0 + \frac{1}{2} + \frac{1}{8} = 1 + \frac{1}{2} + \frac{1}{8} = \frac{13}{8}$ . Thus, $I = \frac{13}{8} \times \frac{\pi}{2} = \frac{13\pi}{16}$ .
verify: Comparing with $I = \frac{k\pi}{16}$ , we get $k = 13$ .