If ∫₀^π 5ᶜᵒˢˣ(1 + cos x cos 3x + cos ²x + cos ³x cos 3x)dx/(1 + 5ᶜᵒˢˣ) = kπ/16 , then k is equal to _____________.

Solution

Maths sir · 4 steps

Replace x by π − x and add: the powers of 5 cancel to 1, and only plain cosines are left.

1. King's property. Under x → π - x, cos x → -cos x and cos 3x → -cos 3x, so cos² x and cos x cos 3x do not change. The bracket factors as g(x) = (1+cos² x)(1+cos x cos 3x), so g does not change either.
2. Add the two forms. In the new form the fraction is 5⁻ᶜᵒˢˣ/(1+5⁻ᶜᵒˢˣ) = 1/(1+5ᶜᵒˢˣ). Add it to the original integral. The two fractions sum to 1.

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

3. Integrate each term. Expand the bracket to 1 + cos² x + cos x cos 3x + cos³ x cos 3x. Over [0,π] the first three terms give π, π/2 and 0. Using cos³ x = (3 cos x + cos 3x)/4, the last term is (3 cos x cos 3x + cos² 3x)/4, which integrates to 1/4(0 + π/2) = π/8.
4. So. 2I = π + π/2 + π/8 = 13π/8, so I = 13π/16 and k = 13.

The common slip: 26. Forgetting to halve after adding the two forms gives 13π/8, that is 26 instead of 13.

The idea to keep. When you see pᶠ⁽ˣ⁾/(1+pᶠ⁽ˣ⁾) and f changes sign under x → a+b-x, add the two forms. The fractions sum to 1, as long as the rest of the integrand stays unchanged.

Answer from the NTA final answer key. Solution by PrepWiser.