The integral 24/π∫₀^(√2) (2 - x²)dx/(2 + x²)√(4 + x⁴) is equal to ____________.

Solution

Maths sir · 4 steps

Divide top and bottom by x² and the substitution t = x + 2/x falls straight out.

1. Divide by x squared. For x > 0, dividing numerator and denominator by x² turns the integrand into (2/x² - 1)/(x + 2/x)√(x² + 4/x²). The square root takes one x inside, and the bracket 2 + x² takes the other.
2. Substitute. Put t = x + 2/x, so dt = (1 - 2/x²)dx, which is minus the numerator. Also x² + 4/x² = t² - 4. As x goes from 0⁺ to √2, t goes from ∞ down to 2√2.

   I = -∫_∞^(2√2) dt/t√(t² - 4) = ∫_(2√2)^∞ dt/t√(t² - 4) = 1/2[sec⁻¹ t/2]_(2√2)^∞

3. Evaluate. At ∞, sec⁻¹ tends to π/2. At t = 2√2, sec⁻¹√2 = π/4. So I = 1/2(π/2 - π/4) = π/8.
4. So. 24/π × π/8 = 3, so the answer is 3.

The common slip: 6. Forgetting the 1/2 in ∫ dt/t√(t² - a²) = 1/a sec⁻¹ t/a makes the integral π/4 and gives 6, not 3.

The idea to keep. When you see a ± x² next to √(a² + x⁴), divide by x². The numerator becomes the derivative of x ± a/x, and the root becomes √(t² ∓ 2a).

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