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junior college 2 | H2 Maths
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Need to use integrstion to solve but i need help!
Date Posted: 4 years ago
Views: 1314
u = sin2x
du/dx = 2cos2x
'' dx = 1/2cos2x du ''
sin²2x cos³2x
= sin²2x(cos2x)(cos²2x)
= sin²2x(cos2x)(1 - sin²2x)
= (sin²2x - sin⁴2x)cos2x
= (u² - u⁴)cos²x
∫ sin²2x cos³2x dx
= ∫ (u² - u⁴)cos2x (1/2cos2x) du
= ½ ∫ (u² - u⁴) du
= ½ (⅓u³ - 1/5 u^5) + C
= 1/6 sin³2x - 1/10 sin^5(2x) + C
du/dx = 2cos2x
'' dx = 1/2cos2x du ''
sin²2x cos³2x
= sin²2x(cos2x)(cos²2x)
= sin²2x(cos2x)(1 - sin²2x)
= (sin²2x - sin⁴2x)cos2x
= (u² - u⁴)cos²x
∫ sin²2x cos³2x dx
= ∫ (u² - u⁴)cos2x (1/2cos2x) du
= ½ ∫ (u² - u⁴) du
= ½ (⅓u³ - 1/5 u^5) + C
= 1/6 sin³2x - 1/10 sin^5(2x) + C
Actually there is no need for substitution in this question as the term is directly integrable.
(sin²2x - sin⁴2x)(cos2x)
= sin²2x cos2x - sin⁴2x cos 2x
cos 2x is the derivative of ½sin2x so it's just doing the reverse of chain rule.
Another way to see :
sub u = sin2x, du/dx = 2 cos2x,
sin²2x cos2x - sin⁴2x cos 2x
= ½ u² du/dx - ½ u⁴ du/dx
= 1/6 (3u² du/dx) - 1/10 (5u⁴ du/dx)
Basically implicit differentiation is done and integration just gives the original expression
(sin²2x - sin⁴2x)(cos2x)
= sin²2x cos2x - sin⁴2x cos 2x
cos 2x is the derivative of ½sin2x so it's just doing the reverse of chain rule.
Another way to see :
sub u = sin2x, du/dx = 2 cos2x,
sin²2x cos2x - sin⁴2x cos 2x
= ½ u² du/dx - ½ u⁴ du/dx
= 1/6 (3u² du/dx) - 1/10 (5u⁴ du/dx)
Basically implicit differentiation is done and integration just gives the original expression
See 1 Answer
done
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clear
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Question already stated that you use method of substitution. The tricky part maybe the du/dx.
Date Posted:
4 years ago
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