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Definite Integration of an Exponential-Trigonometric Fraction Using King's Property

Q: Why do all the cosine harmonic terms cos(2nx) vanish when integrated from 0 to π/2?

Because integral from 0 to π/2 of cos(2nx)dx = [(sin(2nx))/2n]_0^(π/2) = (sin(nπ) - 0)/2n = 0 for all positive integers n.

If integral from 0 to π of (5^(cos x)(1 + cos xcos 3x + cos^2 x + cos^3 xcos 3x)dx)/(1 + 5^(cos x)) = kπ/16, then k is equal to

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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 $I = \int_{0}^{π} \frac{5^{\cos x} (1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x)}{1 + 5^{\cos x}} d x = \frac{k π}{16}$ .

02Approach

Notice that under $x \to π - x$ , $\cos (π - x) = - \cos x$ and $\cos 3 (π - x) = - \cos 3 x$ . 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 $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ eliminates the exponential factor $\frac{5^{\cos x}}{1 + 5^{\cos x}}$ .

03Execute

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

04Execute

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

✓Verify

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

Hints that build this answer step by step

Which symmetry property simplifies the non-algebraic exponential factor $5^{\cos x}$ over the interval $[0, π]$ ?
Apply King's property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ because $\cos (π - x) = - \cos x$ , turning $5^{\cos x}$ into $5^{- \cos x}$ .
How does the trigonometric numerator $N (x) = 1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x$ factor?
$(1 + \cos^{2} x) (1 + \cos x \cos 3 x)$
After applying King's property and adding $2 I = \int_{0}^{π} N (x) d x$ , what is the value of the integral $\int_{0}^{π} N (x) d x$ ?
$\frac{13 π}{8}$
Given $2 I = \frac{13 π}{8}$ and $I = \frac{k π}{16}$ , what is the value of $k$ ?
$k = 13$

✓ Source and academic review

Question type: Numerical
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 do all the cosine harmonic terms

\cos (2 n x)

vanish when integrated from

0

π / 2

Because $\int_{0}^{π / 2} \cos (2 n x) d x = {[\frac{\sin (2 n x)}{2 n}]}_{0}^{π / 2} = \frac{\sin (n π) - 0}{2 n} = 0$ 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 $I = \int_{0}^{π} \frac{5^{\cos x} (1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x)}{1 + 5^{\cos x}} d x = \frac{k π}{16}$ .
Approach: Notice that under $x \to π - x$ , $\cos (π - x) = - \cos x$ and $\cos 3 (π - x) = - \cos 3 x$ . 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 $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ eliminates the exponential factor $\frac{5^{\cos x}}{1 + 5^{\cos x}}$ .
Execute: Applying King's rule: $I = \int_{0}^{π} \frac{5^{- \cos x} (1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x)}{1 + 5^{- \cos x}} d x$ . Adding both expressions for $I$ : $2 I = \int_{0}^{π} (1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x) d x$ . Since the integrand $f (x)$ satisfies $f (π - x) = f (x)$ : $2 I = 2 \int_{0}^{π / 2} (1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x) d x$ $⟹ I = \int_{0}^{π / 2} (1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x) d x$ .
Execute: Factor the integrand: $1 + \cos x \cos 3 x + \cos^{2} x + \cos^{3} x \cos 3 x = (1 + \cos^{2} x) (1 + \cos x \cos 3 x)$ . Alternatively, evaluate directly using trigonometric identities: $\cos x \cos 3 x = \frac{1}{2} (\cos 4 x + \cos 2 x)$ , $\cos^{2} x = \frac{1 + \cos 2 x}{2}$ , $\cos^{3} x \cos 3 x = (\frac{\cos 3 x + 3 \cos x}{4}) \cos 3 x = \frac{1}{4} \cos^{2} 3 x + \frac{3}{4} \cos x \cos 3 x = \frac{1 + \cos 6 x}{8} + \frac{3}{8} (\cos 4 x + \cos 2 x)$ . Integrating from $0$ to $\frac{π}{2}$ , all oscillating terms $\cos (2 n x)$ integrate to zero over $[0, π / 2]$ when $n$ is integer except at the boundaries if any, but specifically: $\int_{0}^{π / 2} \cos 2 x d x = 0$ , $\int_{0}^{π / 2} \cos 4 x d x = 0$ , $\int_{0}^{π / 2} \cos 6 x d x = 0$ . Therefore, the integral only picks up the constant terms: $Constant term = 1 + 0 + \frac{1}{2} + \frac{1}{8} = 1 + \frac{1}{2} + \frac{1}{8} = \frac{13}{8}$ . Thus, $I = \frac{13}{8} \times \frac{π}{2} = \frac{13 π}{16}$ .
Verify: Comparing with $I = \frac{k π}{16}$ , we get $k = 13$ .

Definite Integration of an Exponential-Trigonometric Fraction Using King's Property

What feels right?

The value of k is 13.

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

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