J's answer to Rachel's Junior College 2 H2 Maths Singapore question.
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Here's an alternative method for part 2 using integration by parts :
∫ ( √(1 - 4u²) )³ du
= ∫ (1)(1 - 4u²)³⁄² du
= u (1 - 4u²)³⁄² - ∫ u (3/2 (1 - 4u²)¹⁄² (-8u) ) du
= u (1 - 4u²)³⁄² + ∫ 12u² (1 - 4u²)¹⁄² du
= u (1 - 4u²)³⁄² + 3 ∫ 4u² (1 - 4u²)¹⁄² du
= u (1 - 4u²)³⁄² + ⅜u(8u² - 1)(1 - 4u²)¹⁄² - 3/16 cos-¹(2u)
= u(1 - 4u²)¹⁄² (1 - 4u² + 3u² - ⅜) - 3/16 cos-¹(2u)
= u(⅝ - u²)√(1 - 4u²) - 3/16 cos-¹(2u) + C, C is a constant
∫ ( √(1 - 4u²) )³ du
= ∫ (1)(1 - 4u²)³⁄² du
= u (1 - 4u²)³⁄² - ∫ u (3/2 (1 - 4u²)¹⁄² (-8u) ) du
= u (1 - 4u²)³⁄² + ∫ 12u² (1 - 4u²)¹⁄² du
= u (1 - 4u²)³⁄² + 3 ∫ 4u² (1 - 4u²)¹⁄² du
= u (1 - 4u²)³⁄² + ⅜u(8u² - 1)(1 - 4u²)¹⁄² - 3/16 cos-¹(2u)
= u(1 - 4u²)¹⁄² (1 - 4u² + 3u² - ⅜) - 3/16 cos-¹(2u)
= u(⅝ - u²)√(1 - 4u²) - 3/16 cos-¹(2u) + C, C is a constant
Date Posted:
4 years ago
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