The integral 80 ∫₀^(π/4)((sin θ+cos θ)/(9+16 sin 2 θ)) d θ is equal to :

Solution

Maths sir · 4 steps

Put t = sin θ - cos θ: its derivative is the numerator, and sin 2θ = 1 - t².

1. Choose the substitution. Let t = sin θ - cos θ, so dt = (cos θ + sin θ) dθ. Squaring gives t² = 1 - sin 2θ, so sin 2θ = 1 - t². The denominator becomes 9 + 16(1 - t²) = 25 - 16t².
2. Change the limits. At θ = 0, t = 0 - 1 = -1. At θ = π/4, t = 1/√2 - 1/√2 = 0.

   80∫₋₁⁰dt/(25 - 16t²) = 80 × 1/40[ln|(5 + 4t)/(5 - 4t)|]₋₁⁰

3. Evaluate. This uses ∫ dx/(a² - x²) = 1/2a ln|(a + x)/(a - x)| with x = 4t and a = 5. The extra 1/4 from dx = 4 dt makes the factor 1/4 × 1/10 = 1/40. At t = -1 the fraction is 1/9, so the bracket is ln 1 - ln 1/9 = ln 9.
4. So. The value is 80 × 1/40 × ln 9 = 2 ln 9 = 4 log 3, which is option (b).

The idea to keep. When sin θ + cos θ is on top, put t = sin θ - cos θ so sin 2θ = 1 - t². When cos θ - sin θ is on top, put t = sin θ + cos θ so sin 2θ = t² - 1.

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