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junior college 2 | H2 Maths
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I need help with the first part!
Date Posted: 4 years ago
Views: 841
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I have used the double angle formula for sin and cos several times in the integration and simplification.
Date Posted:
4 years ago
Checking with software suggests that the -1/16 cos-¹(2u) should be 1/16 sin-¹(2u) instead, but your working looks ok to me. Not sure what went amiss
Later I check once again. Could be the angle range or some misintegration.
Doesn't seem to be. I think I know already, it's got to do with the constant.
The two expressions give the same result when differentiated. Should be ok since d/dx sin-¹(x) = 1/√(1-x²) = -d/dx cos-¹(x)
The two expressions give the same result when differentiated. Should be ok since d/dx sin-¹(x) = 1/√(1-x²) = -d/dx cos-¹(x)
Later I try and inspect the equation using the similar looking 2u = sin x
Apparently using 2u = sin x did make the result 1/16 times sin-1 (2u) + the rest.
Yea I tried it too. Works
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I used an alternative substitution 2u = sin x for this.
Date Posted:
4 years ago
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
done
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Part two
Date Posted:
4 years ago
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