Let f(x)=∫ dx/(3+4x²)√(4-3x²),|x|<2/√3. If f(0)=0
and f(1)=1/αβ tan⁻¹(α/β), α,β>0, then α²+β² is equal to ____________.

Solution

Maths sir · 5 steps

Put x = 1/t, then let u² = 4t² − 3: the square root cancels and a plain arctan is left.

1. First substitution. For x > 0 put x = 1/t, so dx = -dt/t². Then 3 + 4x² = (3t²+4)/t² and √(4-3x²) = √(4t²-3)/t, so the integral becomes -∫ (t dt)/(3t²+4)√(4t²-3).
2. Remove the root. Put u² = 4t² - 3, so t dt = (u du)/4 and 3t² + 4 = (3u²+25)/4. The u cancels and you are left with -∫ du/(3u²+25) = -1/5√3 tan⁻¹(√3 u)/5 + C. Since u = √(4-3x²)/x, this gives:

   f(x) = -1/5√3 tan⁻¹((√3 √(4-3x²))/5x) + C

3. Fix the constant. As x → 0⁺ the argument goes to +∞, so the arctan tends to π/2. Then f(0) = 0 gives C = π/10√3.
4. Find f(1). f(1) = 1/5√3(π/2 - tan⁻¹√3/5) = 1/5√3 tan⁻¹ 5/√3, using π/2 - tan⁻¹ θ = tan⁻¹ 1/θ for θ > 0.
5. So. Matching the given form, αβ = 5√3 and α/β = 5/√3, so α = 5 and β = √3. Then α² + β² = 25 + 3 = 28.

The idea to keep. When you see a quadratic times the square root of a quadratic in the denominator, put x = 1/t and then set u² equal to the expression under the root. It reduces to ∫ du/(pu²+q).

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