If ∫ sin⁻¹(√(x/(1+x)))dx = A(x)tan⁻¹(√x) + B(x) + C,
where C is a constant of integration, then the ordered pair (A(x), B(x)) can be :

Solution

Maths sir · 4 steps

Turn the inverse sine into tan⁻¹√x first, then put x = t² and integrate by parts.

1. Simplify the integrand. Let θ = sin⁻¹√(x/(1+x)). Then sin θ = √x/√(1+x) and cos θ = 1/√(1+x), so tan θ = √x. The integral becomes ∫ tan⁻¹√x dx.
2. Substitute. Put x = t², so dx = 2t dt. The integral becomes ∫ 2t tan⁻¹ t dt. Integrate by parts, taking tan⁻¹ t as the first function.

   ∫ 2t tan⁻¹ t dt = t²tan⁻¹ t - ∫ t²/(1+t²) dt = t²tan⁻¹ t - t + tan⁻¹ t + C

3. Back to x. Use t²/(1+t²) = 1 - 1/(1+t²), which gives the last two terms above. Grouping and putting t = √x gives (x+1)tan⁻¹√x - √x + C.
4. So. A(x) = x + 1 and B(x) = -√x. The answer is (a).

The idea to keep. An inverse trig function of a messy algebraic expression often reduces to a simpler one. Draw the right triangle first, then integrate by parts with the inverse trig function as the first function.

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