For 0 < a < 1, the value of the integral ∫0^π (d x)/(1-2 a cos x+a²) is :

Solution

Maths sir · 4 steps

Whenever you see a + b cos x in the denominator, put t = tan x/2 and it becomes a standard arctan.

1. Half-angle substitution. Let t = tan x/2, so cos x = (1-t²)/(1+t²) and dx = (2 dt)/(1+t²). As x goes from 0 to π, t goes from 0 to ∞.
2. Clean the denominator. Multiply the denominator by 1+t²: (1+a²)(1+t²) - 2a(1-t²) = (1-a)² + (1+a)² t². The factor 1+t² cancels with the one in dx.

   I = ∫₀^∞ (2 dt)/((1-a)² + (1+a)² t²) = 2/(1+a)(1-a)[tan⁻¹(1+a)t/(1-a)]₀^∞

3. Evaluate the limits. Since 0 < a < 1, the factor (1+a)/(1-a) is positive, so the arctan runs from 0 to π/2. That gives I = 2/(1-a²) × π/2.
4. So. I = π/(1-a²), which is option (c). As a check, when a → 1 the integrand blows up near x = 0, and only this option blows up too.

The idea to keep. For ∫ dx/(p + q cos x), use t = tan x/2: the denominator turns into a quadratic in t and the answer is an arctan. Always convert the limits along with the variable.

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