Class 12 Maths · Integrals · Very Long Answer #112

📋 CBSE Stepwise Marking Scheme

• [1 mark] For 0<x<1 set u=√x, dx=2u du. The square-root ratio equals (1−u)/√(1−u²).
• [1 mark] Thus I=2∫(u−u²)/√(1−u²)du.
• [1 mark] The first term integrates to −2√(1−u²). Rewrite u²=1−(1−u²); the second term becomes −2arcsin u+2∫√(1−u²)du.
• [1 mark] Use ∫√(1−u²)du=[u√(1−u²)+arcsin u]/2. Combining gives −2√(1−u²)−arcsin u+u√(1−u²). Replace −arcsin u by arccos u up to a constant and substitute u=√x.
• [1 mark] Therefore the required integral is \(\sqrt{x \left(1 - x\right)} - 2 \sqrt{1 - x} + \operatorname{acos}{\left(\sqrt{x} \right)}\) + C.

Master Chapter 7: Integrals