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

Solution

Maths sir · 4 steps

Sine and cosine together in the denominator? Put t = tan x/2 and it becomes a plain quadratic.

1. Substitute. Let t = tan x/2. Then sin x = 2t/(1+t²), cos x = (1-t²)/(1+t²) and dx = (2 dt)/(1+t²). The limits x = 0 and x = π/2 become t = 0 and t = 1.
2. Clean the denominator. Multiply the denominator by 1+t²: 3(1+t²) + 4t + 1 - t² = 2t² + 4t + 4. The 1+t² from dx cancels, and so does the factor 2.

   ∫₀¹ (2 dt)/(2t²+4t+4) = ∫₀¹ dt/((t+1)²+1)

3. Integrate. This is the standard form, so the integral is [tan⁻¹(t+1)]₀¹ = tan⁻¹(2) - tan⁻¹(1).
4. So. Since tan⁻¹(1) = π/4, the integral equals tan⁻¹(2) - π/4, which is option (b).

The idea to keep. For ∫ dx/(a + b sin x + c cos x), use t = tan x/2, then complete the square. Always change the limits along with the variable.

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