JEE Main 2023 · 24 January · evening · Single correct · +4 / −1
∫_(3√2/4)^(3√3/4) 48/√(9 - 4x²)dx is equal to :
- (a) π/2
- (b) π/3
- (c) π/6
- (d) 2π
Answer and solution
Answer (d) 2π
Solution
Maths sir · 3 steps
Pull the 4 out of the root so it looks like √(a² - u²), and the integral becomes an inverse sine.
- Spot the form. Write 9 - 4x² = 3² - (2x)². With u = 2x, du = 2 dx, and ∫ du/√(a² - u²) = sin⁻¹ u/a.
∫ 48/√(9 - 4x²) dx = 48 × 1/2 sin⁻¹ 2x/3 = 24 sin⁻¹ 2x/3
- Put in limits. At x = 3√3/4, 2x/3 = √3/2, so sin⁻¹ gives π/3. At x = 3√2/4, 2x/3 = 1/√2, so sin⁻¹ gives π/4.
- So. The integral is 24(π/3 - π/4) = 24 × π/12 = 2π, which is option (d).
The idea to keep. For ∫ dx/√(a² - b²x²), the answer is 1/b sin⁻¹ bx/a. Do not forget that 1/b, and pick limits that make the inverse sine a standard angle.
Answer from the NTA final answer key. Solution by PrepWiser.