The integral ∫₀^π ((x+3) sin x)/(1+3 cos ² x) d x is equal to

Solution

Maths sir · 4 steps

Split off the x part, then use x → π - x so it turns into the easy integral again.

1. Split the integral. Write I = I₁ + 3J, where I₁ = ∫₀^π (x sin x)/(1+3 cos² x) dx and J = ∫₀^π (sin x)/(1+3 cos² x) dx.
2. Remove the x. Under x → π - x, sin x and cos² x do not change. Adding I₁ to its new form gives 2I₁ = π J, so I₁ = π/2J.

   J = ∫₋₁¹ dt/(1+3t²) = 2/√3 tan⁻¹√3 = 2/√3 × π/3 = 2π/3√3

3. Evaluate J. Put t = cos x, so dt = -sin x dx. The limits 0 to π become 1 to -1, and the minus sign flips them back to -1 to 1, which gives the line above.
4. So. I = π/2 × 2π/3√3 + 3 × 2π/3√3 = (π² + 6π)/3√3 = π/3√3(π+6), which is option (b).

The idea to keep. If the rest of the integrand is unchanged by x → a + b - x, then ∫ₐᵇ x f(x) dx = (a+b)/2∫ₐᵇ f(x) dx. Use this to get rid of a stray x every time.

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