Question

If $\int \limits_0^\pi \frac{5^{\cos x}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right) d x}{1+5^{\cos x}}=\frac{ k \pi}{16}$, then $k$ is equal to

Answer 1

Correct Answer: 26

The correct answer is 26.

\(I=\int_{0}^{\pi}\frac{5^{cosx}(1+cosxcos3x+cos^{2}x+cos^{3}xcos3x)}{1+5^{cosx}}dx\)

\(I=\int_{0}^{\pi}\frac{5^{-cosx}(1+cosxcos3x+cos^{2}x+cos^{3}xcos3x)}{1+5^{-cosx}}dx\)

\(2I=\int_{0}^{\pi}(1+cosxcos3x+cos^{2}x+cos^{3}xcos3x)dx\)

\(\not{2}I=\not{2}\int_{0}^{\frac{\pi}{2}}(1+cosxcos3x+cos^{2}x+cos^{3}xcos3x)dx\)

\(I=\int_{0}^{\frac{\pi}{2}}(1+sinx(-sin3x)+sin^{2}x-sin^{3}xsin3x)dx\)

\(2I=\int_{0}^{\frac{\pi}{2}}(3+cos4x+cos^{3}xcos3x-sin^{3}xsin3x)dx\)

\(2I=\int_{0}^{\frac{\pi}{2}}3+cos4x+(\frac{cos3x+3cosx}{4})cos3x-sin3x(\frac{3sinx-sin3x}{4})dx\)

\(2I=\int_{0}^{\frac{\pi}{2}}(3+cos4x+(\frac{1}{4})+\frac{3}{4}cos4x)dx\)

\(2I=\frac{13}{4}\times\frac{\pi }{2}+\frac{7}{4}(\frac{sin4x}{4})_{0}^{\frac{\pi }{2}}\Rightarrow I=\frac{13\pi }{16}\)

Answer 2

1. Provided Integral: 

I = ∫₀^π (5 cos x (1 + cos x cos 3x + cos² x + cos³ x cos 3x) / (1 + 5 cos x)) dx

2. Symmetry Property:

Consider two integrals:

I = ∫₀^π 5 cos x f(x) dx

I = ∫₀^π 5 (-cos x) f(π - x) dx

Adding these two integrals:

2I = ∫₀^π (1 + cos x cos 3x + cos² x + cos³ x cos 3x) dx

3. Using Symmetry (x → π - x):

2I = 2 ∫₀^π/2 (1 + cos x cos 3x + cos² x + cos³ x cos 3x) dx

4. Simplification Using Trigonometric Identities:

2I = ∫₀^π/2 (3 + cos 4x + (cos x + 3 cos x)/4 + (3 sin x - sin 3x)/4) dx

5. Evaluation of Each Term:

2I = (13/4) · (π/2) + (π/4) - (7/4) · 0 = 13π/16

6. Final Answer:

The value of k is: 13.