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junior college 1 | H2 Maths
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Date Posted: 2 years ago
Views: 498
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First photo of my two-photo workings
I’m not entirely sure of this explanation, but you can take it cautiously.
If a user called J comes in to read this, he might be able to offer a better explanation.
I’m not entirely sure of this explanation, but you can take it cautiously.
If a user called J comes in to read this, he might be able to offer a better explanation.
Date Posted:
2 years ago
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Second photo of my two-photo workings
I use the first-quadrant angle range for simplicity, because later on there will be a problem with the signage of cos z in my workings (will it be positive or negative or both?) if I allow z to take on two quadrants from 0 radian to pi radians
Supposing we allow cos z to be negative (and therefore the final answer will contain a +- in front of the square root).
I have checked that if we were to re-differentiate the version with a negative in front of the square root, then the resulting derivative will be the negative of our original question.
I use the first-quadrant angle range for simplicity, because later on there will be a problem with the signage of cos z in my workings (will it be positive or negative or both?) if I allow z to take on two quadrants from 0 radian to pi radians
Supposing we allow cos z to be negative (and therefore the final answer will contain a +- in front of the square root).
I have checked that if we were to re-differentiate the version with a negative in front of the square root, then the resulting derivative will be the negative of our original question.
Date Posted:
2 years ago
https://www.whitman.edu/mathematics/calculus_late/calculus_late_10_Techniques_of_Integration.pdf
See the explanation under section 10.2.
Since z = arcsin (x/3),
Using the definition of arcsine means −π/2 ≤ z ≤ π/2 , so for this domain, cos z will be ≥ 0 .
So no negative sign is needed.
See the explanation under section 10.2.
Since z = arcsin (x/3),
Using the definition of arcsine means −π/2 ≤ z ≤ π/2 , so for this domain, cos z will be ≥ 0 .
So no negative sign is needed.
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If you have guessed correctly, the similar-looking substitution x = 3 cos z works as well!
Date Posted:
2 years ago
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