The value of ∫₀^π (eᶜᵒˢˣ sin x)/(1 + cos ²x)(eᶜᵒˢˣ + e⁻ᶜᵒˢˣ)dx is equal to:

Solution

Maths sir · 4 steps

Replace x by π - x and add: the exponentials sum to the denominator and cancel.

1. King's property. Under x → π - x, cos x becomes -cos x and sin x stays the same. The factor 1 + cos² x and the sum eᶜᵒˢˣ + e⁻ᶜᵒˢˣ also stay the same. Only the numerator changes, from eᶜᵒˢˣ to e⁻ᶜᵒˢˣ.
2. Add both forms. Adding the original integral to the new one puts eᶜᵒˢˣ + e⁻ᶜᵒˢˣ in the numerator. It cancels with the same factor in the denominator.

   2I = ∫₀^π (sin x)/(1 + cos² x) dx

3. Substitute. Put t = cos x, so dt = -sin x dx, and the limits 0 to π become 1 to -1. Then 2I = ∫₋₁¹ dt/(1+t²) = tan⁻¹(1) - tan⁻¹(-1) = π/4 + π/4 = π/2.
4. So. I = π/4, which is option (c).

The idea to keep. When the integrand has eᶠ over eᶠ + e⁻ᶠ and x → a + b - x flips the sign of f, add the two forms. The awkward fraction becomes 1.

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